# How do you find the volume of the solid obtained by rotating the region bounded by the curves y= x^2 - 4  and y= 3x and x=0 about the y axis?

Sep 28, 2015

$136 \pi$

#### Explanation:

Let's find intersection:

$x = \frac{y}{3} \implies y = {\left(\frac{y}{3}\right)}^{2} - 4$

${y}^{2} / 9 - y - 4 = 0$

${y}^{2} - 9 y - 36 = 0$
${y}^{2} - 12 y + 3 y - 36 = y \left(y - 12\right) + 3 \left(y - 12\right) = 0$
$\left(y - 12\right) \left(y + 3\right) = 0 \iff y = 12 \vee y = - 3$

$y = {x}^{2} - 4 \implies x = \sqrt{y + 4}$
$y = 3 x \implies x = \frac{y}{3}$

$V = \pi \left({\int}_{-} {4}^{12} {\left(\sqrt{y + 4}\right)}^{2} \mathrm{dy} - {\int}_{0}^{12} {\left(\frac{y}{3}\right)}^{2} \mathrm{dy}\right)$

$V = \pi \left({\int}_{-} {4}^{12} \left(y + 4\right) \mathrm{dy} - {\int}_{0}^{12} {y}^{2} / 9 \mathrm{dy}\right)$

${V}_{1} = \pi \left[{y}^{2} / 2 + 4 y\right] {|}_{-} {4}^{12}$

${V}_{1} = \pi \left(144 + 48 - 8 + 16\right) = 200 \pi$

${V}_{2} = \pi \left[{y}^{3} / 27\right] {|}_{0}^{12}$

${V}_{2} = 64 \pi$

$V = 200 \pi - 64 \pi = 136 \pi$