2022 H2 Mathematics Paper 2 Question 5

Definite Integrals: Areas and Volumes

Answers

(a)

13πh2(3r−h).{\frac{1}{3}\pi h^2 (3r-h).}

(b)

p=3.{p = 3.}

(c)

846π cm3.{846 \pi \textrm{ cm}^3.}

Full solutions

(a)

Volume=π∫r−hrx2  dy=π∫r−hrr2−y2  dy=π[r2y−13y3]r−hr=π(r3−13r3−(r2(r−h)−13(r−h)3))=π(23r3−r3+r2h+13(r3−3r2h+3rh2−h3))=π(−13r3+r2h+13r3−r2h+rh2−13h3)=π(rh2−13h3)=13πh2(3r−h)  ■\begin{align*} &\textrm{Volume} \\ &= \pi \int_{r-h}^r x^2 \; \mathrm{d}y \\ &= \pi \int_{r-h}^r r^2 - y^2 \; \mathrm{d}y \\ &= \pi \left[ r^2y - \frac{1}{3} y^3 \right]_{r-h}^r \\ &= \pi \left( r^3 - \frac{1}{3} r^3 - \left( r^2(r-h) - \frac{1}{3}(r-h)^3 \right)\right) \\ &= \pi \left( \frac{2}{3}r^3 - r^3 + r^2h + \frac{1}{3}\left(r^3 - 3r^2h + 3rh^2 - h^3 \right)\right) \\ &= \pi \left( -\frac{1}{3}r^3 + r^2h + \frac{1}{3}r^3 - r^2h + rh^2 - \frac{1}{3}h^3 \right) \\ &= \pi \left( rh^2 - \frac{1}{3}h^3 \right) \\ &= \frac{1}{3}\pi h^2 (3r-h) \; \blacksquare \end{align*}

(b)

Volume of caps=13πp2(3(15)−p)+13π(3p)2(3(15)−3p)=13πp2(45−p+9(45−3p))=13πp2(450−28p)\begin{align*} & \textrm{Volume of caps} \\ &= \frac{1}{3}\pi p^2 \left( 3(15) - p \right) + \frac{1}{3}\pi (3p)^2 \left( 3(15) - 3p \right) \\ &= \frac{1}{3} \pi p^2 \left( 45 - p + 9(45-3p) \right) \\ &= \frac{1}{3} \pi p^2 \left( 450 - 28 p \right) \end{align*}
13πp2(450−28p)=43π(15)3−3402π450p2−28p3=329428p3−450p2+3294=014p3−225p2+1647=0  ■\begin{gather*} \frac{1}{3} \pi p^2 \left( 450 - 28 p \right) = \frac{4}{3}\pi (15)^3 - 3402 \pi \\ 450 p^2 - 28 p^3 = 3294 \\ 28 p^3 - 450 p^2 + 3294 = 0 \\ 14 p^3 - 225 p^2 + 1647 = 0 \; \blacksquare \end{gather*}
From GC,
p=−2.52 (N/A),p=15.6 (N/A) or p=3  ■\begin{align*} p &= -2.52 \textrm{ (N/A)}, \\ p &= 15.6 \textrm{ (N/A)} \textrm{ or } \\ p &= 3 \; \blacksquare \end{align*}

(c)

Volume=13π(3p)2(3r−3p)−13πp2(3r−p)=846π cm3  ■\begin{align*} &\textrm{Volume} \\ &= \frac{1}{3} \pi (3p)^2 (3r - 3p) - \frac{1}{3} \pi p^2 (3r - p) \\ &= 846 \pi \textrm{ cm}^3 \; \blacksquare \end{align*}

Question Commentary

In part (a) of the question, we have to be careful that we want to rotate about the y{y}-axis so we have to use the appropriate volume of integration formula.

The effective way for part (b) is to use the result from (a), replacing h{h} with p{p} and 3p{3p} and subtracting them from the entire sphere.

Thereafter, we should use our graphing calculator to solve the cubic equation, rejecting any negative answers and answers more than 15.{15.}

Part (c) is tricky and involves figuring out how the three conditions provided translates to the picture. It involves shifting the 3p{3p} distance: instead of measuring this length from the bottom as in (b), we now measure a length of 3p{3p} from the top of the sphere. Once we figure this out the calculation is a straightforward application of the formula derived in (a).