Home / Volumes of revolution about the x-axis or y-axis – IBDP Maths analysis and approaches Topic: SL 5.17 HL Paper 2

Volumes of revolution about the x-axis or y-axis – IBDP Maths analysis and approaches Topic: SL 5.17 HL Paper 2

Question.11 [Maximum mark 18]

A function f is defined by \(f(x)=\frac{ke^\frac{x}{2}}{1+e^{x}}\) , where \(x\in R\) , \(x\geq 0 \)and \(k\in R^{+}\)

The region enclosed by the graph of y=(x), the x-axis, the y-axis and the line x = In 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.

Pedro wants to make a small bowl with a volume of \(300 cm^{3}\) 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.

(c) Find

(i) OA

(ii) BC

For design purposes. Pedro investigates how the cross-sectional radius of the bowl changes

(d) (i) By sketching the graph of a suitable derivative off, 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.

▶️Answer/Explanation

(b) a valid algebraic or graphical attempt to find k

\(k^{2}=\frac{300\times 34}{15\Pi }\)
\(k=14.7=\left ( 2\sqrt{\frac{170}{\Pi }}=\sqrt{\frac{680}{\Pi }} \right )\)

c.(i) attempting to find \(OA = f(0)=\frac{k}{2}\)
with \(k=14.7=\left ( 2\sqrt{\frac{170}{\Pi }}=\sqrt{\frac{680}{\Pi }} \right )\)
\(OA=7.36\)

c.(ii) attempting to find \(BC = f(ln16)=\frac{4k}{17}\)
with \(k=14.7=\left ( 2\sqrt{\frac{170}{\Pi }}=\sqrt{\frac{680}{\Pi }} \right )\)
\(BC=3.46\)

d(i)

d(ii)  attempting to find f (1.76… ) 

the cross-sectional radius at this point is \(5.20(\sqrt{\frac{85}{\Pi }})cm\)

Question: [Maximum mark: 5]

The following diagram shows the curve \(\frac{x^{2}}{36}+\frac{(y-4)^{2}}{16}=1,\) where h ≤ y ≤ 4.

The curve from point Q to point B is rotated 360° about the y-axis to form the interior surface of a bowl. The rectangle OPQR, of height h cm, is rotated 360° about the y-axis to form a solid base.
The bowl is assumed to have negligible thickness. Given that the interior volume of the bowl is to be 285 cm3 , determine the height of the base.

▶️Answer/Explanation

Ans:

attempts to express x2 in terms of y

Note: Correct limits are required.

Note: Award M1 for attempting to solve \(36\pi \left ( \frac{h^{3}}{48}-\frac{h^{2}}{4}+\frac{8}{3} \right )=285\) or equivalent for h .
h = 0.7926…
h = 0.793 (cm)

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

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

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.

Question

The region \(A\) is enclosed by the graph of \(y = 2\arcsin (x – 1) – \frac{\pi }{4}\), the \(y\)-axis and the line \(y = \frac{\pi }{4}\).

a.Write down a definite integral to represent the area of \(A\).[4]

b.Calculate the area of \(A\).[2]

▶️Answer/Explanation

Markscheme

METHOD 1

\(2\arcsin (x – 1) – \frac{\pi }{4} = \frac{\pi }{4}\)     (M1)

\(x = 1 + \frac{1}{{\sqrt 2 }}\,\,\,( = 1.707 \ldots )\)     (A1)

\(\int\limits_0^{1 + \frac{1}{{\sqrt 2 }}} {\frac{\pi }{4} – \left( {2\arcsin \left( {x – 1} \right) – \frac{\pi }{4}} \right)dx} \)   M1A1

Note:     Award M1 for an attempt to find the difference between two functions, A1 for all correct.

METHOD 2

when \(x = 0,{\text{ }}y = \frac{{ – 5\pi }}{4}\,\,\,( = – 3.93)\)     A1

\(x = 1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)\)    M1A1

Note:     Award M1 for an attempt to find the inverse function.

\(\int_{\frac{{ – 5\pi }}{4}}^{\frac{\pi }{4}} {\left( {1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)} \right){\text{d}}y} \)     A1

METHOD 3

\(\int_0^{1.38…} {\left( {2\arcsin \left( {x – 1} \right) – \frac{\pi }{4}} \right){\text{d}}x} \left|  +  \right.\int\limits_0^{1.71…} {\frac{\pi }{4}{\text{d}}x – \int\limits_{1.38…}^{1.71…} {\left( {2\arcsin \left( {x – 1} \right) – \frac{\pi }{4}} \right)dx} } \)    M1A1A1A1

Note:     Award M1 for considering the area below the \(x\)-axis and above the \(x\)-axis and A1 for each correct integral.

[4 marks]

a.

\({\text{area}} = 3.30{\text{ (square units)}}\)     A2

[2 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.
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