IB DP Math AI Topic: AHL 5.12 Volumes of revolution IB Style Questions HL Paper 2

Question 4. [Maximum mark: 14]

Charlotte decides to model the shape of a cupcake to calculate its volume.

From rotating a photograph of her cupcake she estimates that its cross-section passes through the points (0 , 3.5), (4 , 6), (6.5 , 4), (7 , 3) and (7.5 , 0), where all units are in centimetres. The cross-section is symmetrical in the x-axis, as shown below:

She models the section from (0 , 3.5) to (4 , 6) as a straight line.

a.   Find the equation of the line passing through these two points. [2]

Charlotte models the section of the cupcake that passes through the points (4 , 6), (6.5 , 4), (7 , 3) and (7.5 , 0) with a quadratic curve.

b.(i)   Find the equation of the least squares regression quadratic curve for these four points.

b. (ii)   By considering the gradient of this curve when x = 4 , explain why it may not be a good model. [3]

Charlotte thinks that a quadratic with a maximum point at (4 , 6) and that passes through the point (7.5 , 0) would be a better fit.

c.   Find the equation of the new model. [4]

Believing this to be a better model for her cupcake, Charlotte finds the volume of revolution about the x-axis to estimate the volume of the cupcake.

d.  (i) Write down an expression for her estimate of the volume as a sum of two integrals.

d.  (ii) Find the value of Charlotte’s estimate. [5]

▶️Answer/Explanation

(a) \(y = \frac{5}{8}x+ \frac{7}{2}(y = 0.625x+ 3.5)\)

(b)(i) \(y= -0.975x^{2}+9.5x-16.7\)

\(y= -0974630x^{2}+9.55919x- 16.6569..)\) 

(ii) gradient of curve is positive at x = 4

(c) let \(y=ax^{2}+bx+c\) differentiating or using \(x=\frac{-b}{2a}8a+b=0\) substituting in the coordinates

\(7.5^{2}a+ 7.5b+c=0    

\(4^{2}a+ 4b+ c=6\)

solve to get \( y =-\frac{24}{49}x^{2}+\frac{192}{49}x-\frac{90}{49}\) OR

\(y = -0.490x^{2}+3.92x-1.84\)

(d)(i) \(\pi \int ^{4}_{0}(\frac{5}{8}x+3.5)^{2}dx+\pi \int ^{7.5}_{4}(-\frac{24}{49}(x-4^{2})+6)^{2}dx\)

(ii) 501cm3(501.189…)

Volumes of Revolution

Rotation About the x-axis

Integration can be used to find the area of a region bounded by a curve whose equation you know. If we want to find the area under the curve y = x2 between x = 0 and x = 5, for example, we simply integrate x2 with limits 0 and 5.
Now imagine that a curve, for example y = x2, is rotated around the x-axis so that a solid is formed. The volume of the shape that is formed can be found using the formula:

Rotation about the y-axis

If the body is rotated about the y-axis rather than the x-axis, then we use the formula:

Volumes of revolution about the  x -axis or y-axis IB Style Questions Examples And Solutions

Question

A function f is defined by 

\(f(x)=\frac{k{e^\frac{x}{2}}}{1+e^{x}}\;where \;x\in \mathbb{R},x\geqslant 0\;and\;k\in \mathbb{R}^+\)

The region enclosed by the graph of y = f (x) , the x-axis, the y-axis and the line x = ln 16 is rotated 360° about the x-axis to form a solid of revolution.

(a) Show that the volume of the solid formed is \(\frac{15k^2\pi}{34} \) cubic units. [6]

Pedro wants to make a small bowl with a volume of 300 cm3 based on the result from part (a). Pedro’s design is shown in the following diagrams.

The vertical height of the bowl, BO, is measured along the x-axis. The radius of the bowl’s top is OA and the radius of the bowl’s base is BC. All lengths are measured in cm.
(b) Find the value of k that satisfies the requirements of Pedro’s design. [2]
(c) Find
(i) OA;
(ii) BC. [4]
For design purposes, Pedro investigates how the cross-sectional radius of the bowl changes.
(d) (i) By sketching the graph of a suitable derivative of f , find where the cross-sectional radius of the bowl is decreasing most rapidly.
      (ii) State the cross-sectional radius of the bowl at this point. [6]

▶️Answer/Explanation

Ans:

(a) $$\begin{eqnarray} \text{req’d vol}_x &=& \pi\int_{0}^{\ln 16}\left[\frac{k\text{e}^{\frac{x}{2}}}{1+\text{e}x}\right]^2\text{d}x \nonumber \\ &=& k^2\pi\int_{0}^{\ln 16}\frac{\text{e}^x}{\left(1+\text{e}^x\right)^2} \nonumber \\ &=& k^2\pi\int_{0}^{\ln 16}\text{e}^x\left(1+\text{e}^x\right)^{-2} \nonumber \\ &=& -k^2\pi\left[\left(1+\text{e}^x\right)^{-1}\right]_{0}^{\ln 16} \nonumber \\ &=& -k^2\pi\left[\left(1+16\right)^{-1}-\left(1+2\right)^{-1}\right] \nonumber \\ &=& -k^2\pi\left(\frac{1}{17}-\frac{1}{2}\right) \nonumber \\ &=& -k^2\pi\left(\frac{2-17}{34}\right) \nonumber \\ &=& \frac{15k^2\pi}{34}. \end{eqnarray}$$ (b) For the volume to be $300\text{cm}^3$, we have $$\begin{eqnarray} 300= \frac{15k^2\pi}{34} \nonumber \\ k=\pm\sqrt{\frac{300\left(34\right)}{15\left(\pi\right)}}, \end{eqnarray}$$ i.e., $k=14.7$. (c)(i) When $x=0$, $f\left(0\right)=\frac{k}{2}=7.36$. Thus, $\text{OA}=7.36\text{ cm}$.
(c)(ii) When $x=\ln 16$, $f\left(\ln 16\right)=\frac{4k}{17}=3.46$. Thus, $\text{BC}=3.46\text{ cm}$.
(d)(i) (graph to be added in later)
(d)(ii) Note that we are finding the minimum point of the derivative graph in part (d)(i). Thus, from the graphing calculator, we have $x=1.76$, i.e., the cross-sectional radius is $f\left(1.76\right)=5.20\text{ cm}$.

Question

The function f is defined by \(f(x) = x\sqrt {9 – {x^2}}  + 2\arcsin \left( {\frac{x}{3}} \right)\).

(a)     Write down the largest possible domain, for each of the two terms of the function, f , and hence state the largest possible domain, D , for f .

(b)     Find the volume generated when the region bounded by the curve y = f(x) , the x-axis, the y-axis and the line x = 2.8 is rotated through \(2\pi \) radians about the x-axis.

(c)     Find \(f'(x)\) in simplified form.

(d)     Hence show that \(\int_{ – p}^p {\frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}} {\text{d}}x = 2p\sqrt {9 – {p^2}}  + 4\arcsin \left( {\frac{p}{3}} \right)\), where \(p \in D\) .

(e)     Find the value of p which maximises the value of the integral in (d).

(f)     (i)     Show that \(f”(x) = \frac{{x(2{x^2} – 25)}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\).

  (ii)     Hence justify that f(x) has a point of inflexion at x = 0 , but not at \(x = \pm \sqrt {\frac{{25}}{2}} \) .

▶️Answer/Explanation

Markscheme

(a)     For \(x\sqrt {9 – {x^2}} \), \( – 3 \leqslant x \leqslant 3\) and for \(2\arcsin \left( {\frac{x}{3}} \right)\), \( – 3 \leqslant x \leqslant 3\)     A1

\( \Rightarrow D{\text{ is }} – 3 \leqslant x \leqslant 3\)     A1

[2 marks]

 

(b)     \(V = \pi \int_0^{2.8} {{{\left( {x\sqrt {9 – {x^2}}  = 2\arcsin \frac{x}{3}} \right)}^2}{\text{d}}x} \)     M1A1

= 181     A1

[3 marks]

 

(c)     \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(9 – {x^2})^{\frac{1}{2}}} – \frac{{{x^2}}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}} + \frac{{\frac{2}{3}}}{{\sqrt {1 – \frac{{{x^2}}}{9}} }}\)     M1A1

\( = {(9 – {x^2})^{\frac{1}{2}}} – \frac{{{x^2}}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}} + \frac{2}{{{{(9 – {x^2})}^{\frac{1}{2}}}}}\)     A1

\( = \frac{{9 – {x^2} – {x^2} + 2}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}}\)     A1

\( = \frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}\)     A1

[5 marks]

 

(d)     \(\int_{ – p}^p {\frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}{\text{d}}x = \left[ {x\sqrt {9 – {x^2}} + 2\arcsin \frac{x}{3}} \right]_{ – p}^p} \)     M1

\( = p\sqrt {9 – {p^2}} + 2\arcsin \frac{p}{3} + p\sqrt {9 – {p^2}}  + 2\arcsin \frac{p}{3}\)     A1

\( = 2p\sqrt {9 – {p^2}} + 4\arcsin \left( {\frac{p}{3}} \right)\)     AG

[2 marks]

 

(e)     \(11 – 2{p^2} = 0\)     M1

\(p = 2.35\,\,\,\,\,\left( {\sqrt {\frac{{11}}{2}} } \right)\)     A1

Note: Award A0 for \(p = \pm 2.35\) .

 

[2 marks]

 

(f)     (i)     \(f”(x) = \frac{{{{(9 – {x^2})}^{\frac{1}{2}}}( – 4x) + x(11 – 2{x^2}){{(9 – {x^2})}^{ – \frac{1}{2}}}}}{{9 – {x^2}}}\)     M1A1

\( = \frac{{ – 4x(9 – {x^2}) + x(11 – 2{x^2})}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\)     A1

\( = \frac{{ – 36x + 4{x^3} + 11x – 2{x^3}}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\)     A1

\( = \frac{{x(2{x^2} – 25)}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\)     AG

 

(ii)     EITHER

When \(0 < x < 3\), \(f”(x) < 0\). When \( – 3 < x < 0\), \(f”(x) > 0\).     A1

OR

\(f”(0) = 0\)     A1

THEN

Hence \(f”(x)\) changes sign through x = 0 , giving a point of inflexion.     R1

EITHER

\(x = \pm \sqrt {\frac{{25}}{2}} \) is outside the domain of f.     R1

OR

\(x = \pm \sqrt {\frac{{25}}{2}} \) is not a root of \(f”(x) = 0\) .     R1

[7 marks]

 

Total [21 marks]

Examiners report

It was disappointing to note that some candidates did not know the domain for arcsin. Most candidates knew what to do in (b) but sometimes the wrong answer was obtained due to the calculator being in the wrong mode. In (c), the differentiation was often disappointing with \(\arcsin \left( {\frac{x}{3}} \right)\) causing problems. In (f)(i), some candidates who failed to do (c) guessed the correct form of \(f'(x)\) (presumably from (d)) and then went on to find \(f”(x)\) correctly. In (f)(ii), the justification of a point of inflexion at x = 0 was sometimes incorrect – for example, some candidates showed simply that \(f'(x)\) is positive on either side of the origin which is not a valid reason.

Question

Let \(f\) be a function defined by \(f(x) = x + 2\cos x\) , \(x \in \left[ {0,{\text{ }}2\pi } \right]\) . The diagram below shows a region \(S\) bound by the graph of \(f\) and the line \(y = x\) .

A and C are the points of intersection of the line \(y = x\) and the graph of \(f\) , and B is the minimum point of \(f\) .

(a)     If A, B and C have x-coordinates \(a\frac{\pi }{2}\), \(b\frac{\pi }{6}\) and \(c\frac{\pi }{2}\), where \(a\) , \(b\), \(c \in \mathbb{N}\) , find the values of \(a\) , \(b\) and \(c\) .

(b)     Find the range of \(f\) .

(c)     Find the equation of the normal to the graph of f at the point C, giving your answer in the form \(y = px + q\) .

(d)     The region \(S\) is rotated through \({2\pi }\) about the x-axis to generate a solid.

  (i)     Write down an integral that represents the volume \(V\) of this solid.

  (ii)     Show that \(V = 6{\pi ^2}\) .

▶️Answer/Explanation

Markscheme

(a)     METHOD 1

using GDC

\(a = 1\), \(b = 5\), \(c = 3\)     A1A2A1

METHOD 2

\(x = x + 2\cos x \Rightarrow \cos x = 0\)

\( \Rightarrow x = \frac{\pi }{2}\), \(\frac{{3\pi }}{2}\) …     M1

\(a = 1\), \(c = 3\)     A1

\(1 – 2\sin x = 0\)     M1

\( \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi }{6}\) or \(\frac{{5\pi }}{6}\)

\(b = 5\)     A1

Note: Final M1A1 is independent of previous work.

[4 marks]

(b)     \(f\left( {\frac{{5\pi }}{6}} \right) = \frac{{5\pi }}{6} – \sqrt 3 \)   (or \(0.886\))     (M1)

\(f(2\pi ) = 2\pi  + 2\) (or \(8.28\))     (M1)

the range is \(\left[ {\frac{{5\pi }}{6} – \sqrt 3 ,{\text{ }}2\pi  + 2} \right]\) (or [\(0.886\), \(8.28\)])     A1

[3 marks]

(c)     \(f'(x) = 1 – 2\sin x\)     (M1)

\(f’\left( {\frac{{3\pi }}{6}} \right) = 3\)     A1

gradient of normal \( = – \frac{1}{3}\)     (M1)

equation of the normal is \(y – \frac{{3\pi }}{2} = – \frac{1}{3}\left( {x – \frac{{3\pi }}{2}} \right)\)     (M1)

\(y = – \frac{1}{3}x + 2\pi \)   (or equivalent decimal values)     A1     N4

[5 marks]

(d)     (i)     \(V = \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{x^2} – {{\left( {x + 2\cos x} \right)}^2}} \right)} {\text{d}}x\)   (or equivalent)     A1A1

Note: Award A1 for limits and A1 for \(\pi \) and integrand.

(ii)     \(V = \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{x^2} – {{\left( {x + 2\cos x} \right)}^2}} \right)} {\text{d}}x\)

\( = – \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {4x\cos x + 4{{\cos }^2}x} \right)} {\text{d}}x\)

using integration by parts     M1

and the identity \(4{\cos ^2}x = 2\cos 2x + 2\) ,     M1

\(V = – \pi \left[ {\left( {4x\sin x + 4\cos x} \right) + \left( {\sin 2x + 2x} \right)} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}\)     A1A1

Note: Award A1 for \({4x\sin x + 4\cos x}\) and A1 for sin \({2x + 2x}\) .

\( = – \pi \left[ {\left( {6\pi \sin \frac{{3\pi }}{2} + 4\cos \frac{{3\pi }}{2} + \sin 3\pi  + 3\pi } \right) – \left( {2\pi sin\frac{\pi }{2} + 4\cos \frac{\pi }{2} + \sin \pi  + \pi } \right)} \right]\)     A1

\( = – \pi \left( { – 6\pi  + 3\pi  – \pi } \right)\)

\( = 6{\pi ^2}\)     AG     N0

Note: Do not accept numerical answers.

[7 marks]

Total [19 marks]

Examiners report

Generally there were many good attempts to this, more difficult, question. A number of students found \(b\) to be equal to 1, rather than 5. In the final part few students could successfully work through the entire integral successfully.

Question

Let \(f(x) = \frac{{a + b{{\text{e}}^x}}}{{a{{\text{e}}^x} + b}}\), where \(0 < b < a\).

(a)     Show that \(f'(x) = \frac{{({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\).

(b)     Hence justify that the graph of f has no local maxima or minima.

(c)     Given that the graph of f has a point of inflexion, find its coordinates.

(d)     Show that the graph of f has exactly two asymptotes.

(e)     Let a = 4 and b =1. Consider the region R enclosed by the graph of \(y = f(x)\), the y-axis and the line with equation \(y = \frac{1}{2}\).

Find the volume V of the solid obtained when R is rotated through \(2\pi \) about the x-axis.

▶️Answer/Explanation

Markscheme

(a)     \(f'(x) = \frac{{b{{\text{e}}^x}(a{{\text{e}}^x} + b) – a{{\text{e}}^x}(a + b{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^2}}}\)     M1A1

\( = \frac{{ab{{\text{e}}^{2x}} + {b^2}{{\text{e}}^x} – {a^2}{{\text{e}}^x} – ab{{\text{e}}^{2x}}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\)     A1

\( = \frac{{({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\)     AG

[3 marks]

 

(b)     EITHER

\(f'(x) = 0 \Rightarrow ({b^2} – {a^2}){{\text{e}}^x} = 0 \Rightarrow b =  \pm a{\text{ or }}{{\text{e}}^x} = 0\)     A1

which is impossible as \(0 < b < a\) and \({{\text{e}}^x} > 0\) for all \(x \in \mathbb{R}\)     R1

OR

\(f'(x) < 0\) for all \(x \in \mathbb{R}\) since \(0 < b < a\) and \({{\text{e}}^x} > 0\) for all \(x \in \mathbb{R}\)     A1R1

OR

\(f'(x)\) cannot be equal to zero because \({{\text{e}}^x}\) is never equal to zero     A1R1

[2 marks]

 

(c)     EITHER

\(f”(x) = \frac{{({b^2} – {a^2}){{\text{e}}^x}{{(a{{\text{e}}^x} + b)}^2} – 2a{{\text{e}}^x}(a{{\text{e}}^x} + b)({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^4}}}\)     M1A1A1

Note: Award A1 for each term in the numerator.

 

\( = \frac{{({b^2} – {a^2}){{\text{e}}^x}(a{{\text{e}}^x} + b – 2a{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^3}}}\)

\( = \frac{{({b^2} – {a^2})(b – a{{\text{e}}^x}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^3}}}\)

OR

\(f'(x) = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 2}}\)

\(f”(x) = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 2}} + ({b^2} – {a^2}){{\text{e}}^x}( – 2a{{\text{e}}^x}){(a{{\text{e}}^x} + b)^{ – 3}}\)     M1A1A1

Note: Award A1 for each term.

 

\( = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 3}}\left( {(a{{\text{e}}^x} + b) – 2a{{\text{e}}^x}} \right)\)

\( = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 3}}(b – a{{\text{e}}^x})\)

THEN

\(f”(x) = 0 \Rightarrow b – a{{\text{e}}^x} = 0 \Rightarrow x = \ln \frac{b}{a}\)     M1A1

\(f\left( {\ln \frac{b}{a}} \right) = \frac{{{a^2} + {b^2}}}{{2ab}}\)     A1

coordinates are \(\left( {\ln \frac{b}{a},\frac{{{a^2} + {b^2}}}{{2ab}}} \right)\)

[6 marks]

 

(d)     \(\mathop {\lim }\limits_{x – \infty } f(x) = \frac{a}{b} \Rightarrow y = \frac{a}{b}\) horizontal asymptote     A1

\(\mathop {\lim }\limits_{x \to  + \infty } f(x) = \frac{b}{a} \Rightarrow y = \frac{b}{a}\) horizontal asymptote     A1

\(0 < b < a \Rightarrow a{{\text{e}}^x} + b > 0\) for all \(x \in \mathbb{R}\) (accept \(a{{\text{e}}^x} + b \ne 0\))

so no vertical asymptotes     R1

Note: Statement on vertical asymptote must be seen for R1.

 

[3 marks]

 

(e)     \(y = \frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}\)

\(y = \frac{1}{2} \Leftrightarrow x = \ln \frac{7}{2}\) (or 1.25 to 3 sf)     (M1)(A1)

\(V = \pi \int_0^{\ln \frac{7}{2}} {\left( {{{\left( {\frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}} \right)}^2} – \frac{1}{4}} \right){\text{d}}x} \)     (M1)A1

\( = 1.09\) (3 sf)     A1     N4

[5 marks]

Total [19 marks]

Examiners report

This question was well attempted by many candidates. In some cases, candidates who skipped other questions still answered, with some success, parts of this question. Part (a) was in general well done but in (b) candidates found difficulty in justifying that f’(x) was non-zero. Performance in part (c) was mixed: it was pleasing to see good levels of algebraic ability of good candidates who successfully answered this question; weaker candidates found the simplification required difficult. There were very few good answers to part (d) which showed the weaknesses of most candidates in dealing with the concept of asymptotes. In part (e) there were a large number of good attempts, with many candidates evaluating correctly the limits of the integral and a smaller number scoring full marks in this part.

Question

A cone has height h and base radius r . Deduce the formula for the volume of this cone by rotating the triangular region, enclosed by the line \(y = h – \frac{h}{r}x\) and the coordinate axes, through \(2\pi \) about the y-axis.

▶️Answer/Explanation

Markscheme

\(x = r – \frac{r}{h}y{\text{ or }}x = \frac{r}{h}(h – y){\text{ (or equivalent)}}\)     (A1)

\(\int {\pi {x^2}{\text{d}}y} \)

\( = \pi \int_0^h {{{\left( {r – \frac{r}{h}y} \right)}^2}{\text{d}}y} \)     M1A1 

Note: Award M1 for \(\int {{x^2}{\text{d}}y} \) and A1 for correct expression.

Accept \(\pi \int_0^h {{{\left( {\frac{r}{h}y – r} \right)}^2}{\text{d}}y{\text{ and }}\pi \int_0^h {{{\left( { \pm \left( {r – \frac{r}{h}x} \right)} \right)}^2}{\text{d}}x} } \)

\( = \pi \int_0^h {\left( {{r^2} – \frac{{2{r^2}}}{h}y + \frac{{{r^2}}}{{{h^2}}}{y^2}} \right){\text{d}}y} \)     A1

Note: Accept substitution method and apply markscheme to corresponding steps.

\( = \pi \left[ {{r^2}y – \frac{{{r^2}{y^2}}}{h} + \frac{{{r^2}{y^3}}}{{3{h^2}}}} \right]_0^h\)     M1A1 

Note: Award M1 for attempted integration of any quadratic trinomial.

\( = \pi \left( {{r^2}h – {r^2}h + \frac{1}{3}{r^2}h} \right)\)     M1A1 

Note: Award M1 for attempted substitution of limits in a trinomial.

\( = \frac{1}{3}\pi {r^2}h\)     A1 

Note: Throughout the question do not penalize missing dx/dy as long as the integrations are done with respect to correct variable.

 

[9 marks]

Examiners report

Most candidates attempted this question using either the formula given in the information booklet or the disk method. However, many were not successful, either because they started off with the incorrect expression or incorrect integration limits or even attempted to integrate the correct expression with respect to the incorrect variable.

Question

Find the volume of the solid formed when the region bounded by the graph of \(y = \sin (x – 1)\), and the lines y = 0 and y = 1 is rotated by \(2\pi \) about the y-axis.

▶️Answer/Explanation

Markscheme

volume \( = \pi \int {{x^2}{\text{d}}y} \)     (M1)

\(x = \arcsin y + 1\)     (M1)(A1)

volume \( = \pi \int_0^1 {{{(\arcsin y + 1)}^2}{\text{d}}y} \)     A1

Note:A1 is for the limits, provided a correct integration of y.

 

\( = 2.608993 \ldots \pi  = 8.20\)     A2     N5

[6 marks]

Examiners report

Although it was recognised that the imprecise nature of the wording of the question caused some difficulties, these were overwhelmingly by candidates who were attempting to rotate around the \(x\)-axis. The majority of students who understood to rotate about the \(y\)-axis had no difficulties in writing the correct integral. Marks lost were for inability to find the correct value of the integral on the GDC (some clearly had the calculator in degrees) and also for poor rounding where the GDC had been used correctly. In the few instances where students seemed confused by the lack of precision in the question, benefit of the doubt was given and full points awarded.

Question

The shaded region S is enclosed between the curve \(y = x + 2\cos x\), for \(0 \leqslant x \leqslant 2\pi \), and the line \(y = x\), as shown in the diagram below.



Find the coordinates of the points where the line meets the curve.[3]

a.

The region \(S\) is rotated by \(2\pi \) about the \(x\)-axis to generate a solid.

(i)     Write down an integral that represents the volume \(V\) of the solid.

(ii)     Find the volume \(V\).[5]

b.
▶️Answer/Explanation

Markscheme

(a)     \(\frac{\pi }{2}(1.57),{\text{ }}\frac{{3\pi }}{2}(4.71)\)     A1A1

hence the coordinates are \(\left( {\frac{\pi }{2},{\text{ }}\frac{\pi }{2}} \right),{\text{ }}\left( {\frac{{3\pi }}{2},{\text{ }}\frac{{3\pi }}{2}} \right)\)     A1

[3 marks]

a.

(i)     \(\pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{x^2} – {{(x + 2\cos x)}^2}} \right){\text{d}}x} \)     A1A1A1

 

Note:     Award A1 for \({x^2} – {(x + 2\cos x)^2}\), A1 for correct limits and A1 for \(\pi \).

 

(ii)     \(6{\pi ^2}{\text{ }}( = 59.2)\)     A2

 

Notes:     Do not award ft from (b)(i).

 

[5 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

The vertical cross-section of a container is shown in the following diagram.

The curved sides of the cross-section are given by the equation \(y = 0.25{x^2} – 16\). The horizontal cross-sections are circular. The depth of the container is \(48\) cm.

If the container is filled with water to a depth of \(h\,{\text{cm}}\), show that the volume, \(V\,{\text{c}}{{\text{m}}^3}\), of the water is given by \(V = 4\pi \left( {\frac{{{h^2}}}{2} + 16h} \right)\).

[3]
a.

The container, initially full of water, begins leaking from a small hole at a rate given by \(\frac{{{\text{d}}V}}{{{\text{d}}t}} =  – \frac{{250\sqrt h }}{{\pi(h + 16)}}\) where \(t\) is measured in seconds.

(i)     Show that \(\frac{{{\text{d}}h}}{{{\text{d}}t}} =  – \frac{{250\sqrt h }}{{4{\pi ^2}{{(h + 16)}^2}}}\).

(ii)     State \(\frac{{{\text{d}}t}}{{{\text{d}}h}}\) and hence show that \(t = \frac{{ – 4{\pi ^2}}}{{250}}\int {\left( {{h^{\frac{3}{2}}} + 32{h^{\frac{1}{2}}} + 256{h^{ – \frac{1}{2}}}} \right){\text{d}}h} \).

(iii)     Find, correct to the nearest minute, the time taken for the container to become empty. (\(60\) seconds = 1 minute)

[10]
b.

Once empty, water is pumped back into the container at a rate of \(8.5\;{\text{c}}{{\text{m}}^3}{{\text{s}}^{ – 1}}\). At the same time, water continues leaking from the container at a rate of \(\frac{{250\sqrt h }}{{\pi (h + 16)}}{\text{c}}{{\text{m}}^3}{{\text{s}}^{ – 1}}\).

Using an appropriate sketch graph, determine the depth at which the water ultimately stabilizes in the container.

[3]
c.
▶️Answer/Explanation

Markscheme

attempting to use \(V = \pi \int_a^b {{x^2}{\text{d}}y} \)     (M1)

attempting to express \({x^2}\) in terms of \(y\) ie \({x^2} = 4(y + 16)\)     (M1)

for \(y = h,{\text{ }}V = 4\pi \int_0^h {y + 16{\text{d}}y} \)     A1

\(V = 4\pi \left( {\frac{{{h^2}}}{2} + 16h} \right)\)     AG

[3 marks]

a.

(i)     METHOD 1

\(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{{{\text{d}}h}}{{{\text{d}}V}} \times \frac{{{\text{d}}V}}{{{\text{d}}t}}\)     (M1)

\(\frac{{{\text{d}}V}}{{{\text{d}}h}} = 4\pi (h + 16)\)     (A1)

\(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{1}{{4\pi (h + 16)}} \times \frac{{ – 250\sqrt h }}{{\pi (h + 16)}}\)     M1A1

Note:     Award M1 for substitution into \(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{{{\text{d}}h}}{{{\text{d}}V}} \times \frac{{{\text{d}}V}}{{{\text{d}}t}}\).

\(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{{250\sqrt h }}{{4{\pi ^2}{{(h + 16)}^2}}}\)     AG

METHOD 2

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = 4\pi (h + 16)\frac{{{\text{d}}h}}{{{\text{d}}t}}\;\;\;\)(implicit differentiation)(M1)

\(\frac{{ – 250\sqrt h }}{{\pi (h + 16)}} = 4\pi (h + 16)\frac{{{\text{d}}h}}{{{\text{d}}t}}\;\;\;\)(or equivalent)     A1

\(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{1}{{4\pi (h + 16)}} \times \frac{{ – 250\sqrt h }}{{\pi (h + 16)}}\)     M1A1

\(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{{250\sqrt h }}{{4{\pi ^2}{{(h + 16)}^2}}}\)     AG

(ii)     \(\frac{{{\text{d}}t}}{{{\text{d}}h}} =  – \frac{{4{\pi ^2}{{(h + 16)}^2}}}{{250\sqrt h }}\)     A1

\(t = \int { – \frac{{4{\pi ^2}{{(h + 16)}^2}}}{{250\sqrt h }}} {\text{d}}h\)     (M1)

\(t = \int { – \frac{{4{\pi ^2}({h^2} + 32h + 256)}}{{250\sqrt h }}} {\text{d}}h\)     A1

\(t = \frac{{ – 4{\pi ^2}}}{{250}}\int {\left( {{h^{\frac{3}{2}}} + 32{h^{\frac{1}{2}}} + 256{h^{ – \frac{1}{{2}}}}} \right){\text{d}}h} \)     AG

(iii)     METHOD 1

\(t = \frac{{ – 4{\pi ^2}}}{{250}}\int_{48}^0 {\left( {{h^{\frac{3}{2}}} + 32{h^{\frac{1}{2}}} + 256{h^{ – \frac{1}{2}}}} \right)} {\text{d}}h\)     (M1)

\(t = 2688.756 \ldots {\text{ (s)}}\)     (A1)

\(45\) minutes (correct to the nearest minute)     A1

METHOD 2

\(t = \frac{{ – 4{\pi ^2}}}{{250}}\left( {\frac{2}{5}{h^{\frac{5}{2}}} + \frac{{64}}{3}{h^{\frac{3}{2}}} + 512{h^{\frac{1}{2}}}} \right) + c\)

when \(t = 0,{\text{ }}h = 48 \Rightarrow c = 2688.756 \ldots \left( {c = \frac{{4{\pi ^2}}}{{250}}\left( {\frac{2}{5} \times {{48}^{\frac{5}{2}}} + \frac{{64}}{3} \times {{48}^{\frac{3}{2}}} + 512 \times {{48}^{\frac{1}{2}}}} \right)} \right)\)     (M1)

when \(h = 0,{\text{ }}t = 2688.756 \ldots \left( {t = \frac{{4{\pi ^2}}}{{250}}\left( {\frac{2}{5} \times {{48}^{\frac{5}{2}}} + \frac{{64}}{3} \times {{48}^{\frac{3}{2}}} + 512 \times {{48}^{\frac{1}{2}}}} \right)} \right){\text{ (s)}}\)     (A1)

45 minutes (correct to the nearest minute)     A1

[10 marks]

b.

EITHER

the depth stabilizes when \(\frac{{{\text{d}}V}}{{{\text{d}}t}} = 0\;\;\;ie\;\;\;8.5 – \frac{{250\sqrt h }}{{\pi (h + 16)}} = 0\)     R1

attempting to solve \(8.5 – \frac{{250\sqrt h }}{{\pi (h + 16)}} = 0\;\;\;{\text{for }}h\)     (M1)

OR

the depth stabilizes when \(\frac{{{\text{d}}h}}{{{\text{d}}t}} = 0\;\;\;ie\;\;\;\frac{1}{{4\pi (h + 16)}}\left( {8.5 – \frac{{250\sqrt h }}{{\pi (h + 16)}}} \right) = 0\)     R1

attempting to solve \(\frac{1}{{4\pi (h + 16)}}\left( {8.5 – \frac{{250\sqrt h }}{{\pi (h + 16)}}} \right) = 0\;\;\;{\text{for }}h\)     (M1)

THEN

\(h = 5.06{\text{ (cm)}}\)     A1

[3 marks]

Total [16 marks]

c.

Examiners report

This question was done reasonably well by a large proportion of candidates. Many candidates however were unable to show the required result in part (a). A number of candidates seemingly did not realize how the container was formed while other candidates attempted to fudge the result.

a.

Part (b) was quite well done. In part (b) (i), most candidates were able to correctly calculate \(\frac{{{\text{d}}V}}{{{\text{d}}h}}\) and correctly apply a related rates expression to show the given result. Some candidates however made a sign error when stating \(\frac{{{\text{d}}V}}{{{\text{d}}t}}\). A large number of candidates successfully answered part (b) (ii). In part (b) (iii), successful candidates either set up and calculated an appropriate definite integral or antidifferentiated and found that \(t = C\) when \(h = 0\).

b.

In part (c), a pleasing number of candidates realized that the water depth stabilized when either \(\frac{{{\text{d}}V}}{{{\text{d}}t}} = 0\) or \(\frac{{{\text{d}}h}}{{{\text{d}}t}} = 0\), sketched an appropriate graph and found the correct value of \(h\). Some candidates misinterpreted the situation and attempted to find the coordinates of the local minimum of their graph.

c.

Question

The region \(R\) is enclosed by the graph of \(y = {e^{ – {x^2}}}\), the \(x\)-axis and the lines \(x =  – 1\) and \(x = 1\).

Find the volume of the solid of revolution that is formed when \(R\) is rotated through \(2\pi \) about the \(x\)-axis.

▶️Answer/Explanation

Markscheme

\(\int_{ – 1}^1 {\pi {{\left( {{{\text{e}}^{ – {x^2}}}} \right)}^2}{\text{d}}x} \;\;\;\left( {\int_{ – 1}^1 {\pi {{\text{e}}^{ – 2{x^2}}}{\text{d}}x} \;\;\;{\text{or}}\;\;\;\int_0^1 {2\pi {{\text{e}}^{ – 2{x^2}}}{\text{d}}x} } \right)\)     (M1)(A1)(A1)

Note:     Award M1 for integral involving the function given; A1 for correct limits; A1 for \(\pi \) and \({{{\left( {{{\text{e}}^{ – {x^2}}}} \right)}^2}}\)

\( = 3.758249 \ldots  = 3.76\)     A1

[4 marks]

Examiners report

Most candidates answered this question correctly. Those candidates who attempted to manipulate the function or attempt an integration wasted time and obtained 3/4 marks. The most common errors were an extra factor ‘2’ and a fourth power when attempting to square the function. Many candidates wrote down the correct expression but not all were able to use their calculator correctly.

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