2017 H2 Mathematics Paper 2 Question 4

Definite Integrals: Areas and Volumes

Answers

(a)

15.1875 units2{15.1875 \textrm{ units}^2}
(bi)
π2a(a−1) units3{\frac{\pi}{2a(a-1)} \textrm{ units}^3}
(bii)
b=1+a2−a+12{b = \frac{1 + \sqrt{a^2 - a + 1}}{2}}

Full solutions

(a)

Using a GC, the two graphs intersect at x=1{x=1} and x=5.5{x=5.5}
Area of the plate=∫15.5(x−12−(x2−6x+5))  dx=15.1875 units2  ■\begin{align*} & \textrm{Area of the plate} \\ & = \int_{1}^{5.5} \left( \frac{x-1}{2} - (x^2 - 6 x + 5) \right) \; \mathrm{d}x \\ & = 15.1875 \textrm{ units}^2 \; \blacksquare \end{align*}
(bi)
Volume of container=π∫01x2  dy=π∫01(ya−y2)2  dy=π∫01y(a−y2)2  dy=−π2∫01−2y(a−y2)−2  dy=−π2[(a−y2)−1−1]01=π2(1a−1−1a)=π(a−(a−1))2a(a−1)=π2a(a−1) units3  ■\begin{align*} & \textrm{Volume of container} \\ & = \pi \int_0^{1} x^2 \; \mathrm{d}y \\ & = \pi \int_0^{1} \left( \frac{\sqrt{y}}{a-y^2} \right)^2 \; \mathrm{d}y \\ & = \pi \int_0^{1} \frac{y}{(a-y^2)^2} \; \mathrm{d}y \\ & = - \frac{\pi}{2} \int_0^{1} -2y (a-y^2)^{-2} \; \mathrm{d}y \\ & = - \frac{\pi}{2} \left[ \frac{(a-y^2)^{-1}}{-1} \right]_0^1 \\ & = \frac{\pi}{2} \left( \frac{1}{a-1} - \frac{1}{a} \right) \\ & = \frac{\pi(a-(a-1))}{2a(a-1)} \\ & = \frac{\pi}{2a(a-1)} \textrm{ units}^3 \; \blacksquare \\ \end{align*}
(bii)
π2b(b−1)=4(π2a(a−1))4b(b−1)=a(a−1)4b2−4b−a(a−1)=0\begin{gather*} \frac{\pi}{2b(b-1)} = 4 \left( \frac{\pi}{2a(a-1)} \right) \\ 4b(b-1) = a(a-1) \\ 4b^2 - 4b - a(a-1) = 0 \\ \end{gather*}
b=4±16−4(4)(−a(a−1))8=4±16+16a2−16a8=4±4a2−a+18=1±a2−a+12\begin{align*} b &= \frac{4 \pm \sqrt{16 - 4(4)(-a(a-1))}}{8} \\ &= \frac{4 \pm \sqrt{16 + 16a^2 - 16a}}{8} \\ &= \frac{4 \pm 4 \sqrt{a^2 - a + 1}}{8} \\ &= \frac{1 \pm \sqrt{a^2 - a + 1}}{2} \\ \end{align*}
Since a,b>1,{a, b > 1,}
b=1+a2−a+12  ■b = \frac{1 + \sqrt{a^2 - a + 1}}{2} \; \blacksquare