Question
A function is defined by \(f(x) = {x^2} + 2,{\text{ }}x \ge 0\). A region \(R\) is enclosed by \(y = f(x)\),the \(y\)-axis and the line \(y = 4\).
a.(i) Express the area of the region \(R\) as an integral with respect to \(y\).
(ii) Determine the area of \(R\), giving your answer correct to four significant figures.[3]
b.Find the exact volume generated when the region \(R\) is rotated through \(2\pi \) radians about the \(y\)-axis.[3]
▶️Answer/Explanation
Markscheme
(i) \({\text{area}} = \int_2^4 {\sqrt {y – 2} {\text{d}}y} \) M1A1
(ii) \( = 1.886{\text{ (4 sf only)}}\) A1
Note: Award M0A0A1 for finding \(1.886\) from \(\int_0^{\sqrt 2 } {4 – f(x){\text{d}}x} \).
Award A1FT for a 4sf answer obtained from an integral involving \(x\).
[3 marks]
\({\text{volume}} = \pi \int_2^4 {(y – 2){\text{d}}y} \) (M1)
Note: Award M1 for the correct integral with incorrect limits.
\( = \pi \left[ {\frac{{{y^2}}}{2} – 2y} \right]_2^4\) (A1)
\( = 2\pi {\text{ (exact only)}}\) A1
[3 marks]
Total [6 marks]
Examiners report
[N/A]
[N/A]
Question
The following graph shows the two parts of the curve defined by the equation \({x^2}y = 5 – {y^4}\), and the normal to the curve at the point P(2 , 1).
a.Show that there are exactly two points on the curve where the gradient is zero.[7]
b.Find the equation of the normal to the curve at the point P.[5]
c.The normal at P cuts the curve again at the point Q. Find the \(x\)-coordinate of Q.[3]
d.The shaded region is rotated by 2\(\pi \) about the \(y\)-axis. Find the volume of the solid formed.[7]
▶️Answer/Explanation
Markscheme
differentiating implicitly: M1
\(2xy + {x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) A1A1
Note: Award A1 for each side.
if \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) then either \(x = 0\) or \(y = 0\) M1A1
\(x = 0 \Rightarrow \) two solutions for \(y\left( {y = \pm \sqrt[4]{5}} \right)\) R1
\(y = 0\) not possible (as 0 ≠ 5) R1
hence exactly two points AG
Note: For a solution that only refers to the graph giving two solutions at \(x = 0\) and no solutions for \(y = 0\) award R1 only.
[7 marks]
at (2, 1) \(4 + 4\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 4\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{1}{2}\) (A1)
gradient of normal is 2 M1
1 = 4 + c (M1)
equation of normal is \(y = 2x – 3\) A1
[5 marks]
substituting (M1)
\({x^2}\left( {2x – 3} \right) = 5 – {\left( {2x – 3} \right)^4}\) or \({\left( {\frac{{y + 3}}{2}} \right)^2}\,y = 5 – {y^4}\) (A1)
\(x = 0.724\) A1
[3 marks]
recognition of two volumes (M1)
volume \(1 = \pi \int_1^{\sqrt[4]{5}} {\frac{{5 – {y^4}}}{y}} {\text{d}}y\left( { = 101\pi = 3.178 \ldots } \right)\) M1A1A1
Note: Award M1 for attempt to use \(\pi \int {{x^2}} {\text{d}}y\), A1 for limits, A1 for \({\frac{{5 – {y^4}}}{y}}\) Condone omission of \(\pi \) at this stage.
volume 2
EITHER
\( = \frac{1}{3}\pi \times {2^2} \times 4\left( { = 16.75 \ldots } \right)\) (M1)(A1)
OR
\( = \pi \int_{ – 3}^1 {{{\left( {\frac{{y + 3}}{2}} \right)}^2}} {\text{d}}y\left( { = \frac{{16\pi }}{3} = 16.75 \ldots } \right)\) (M1)(A1)
THEN
total volume = 19.9 A1
[7 marks]
Examiners report
[N/A]
[N/A]
[N/A]
[N/A]