Quiz 6.1:
Question:
Find the volume generated in revolving the region enclosed by \( x = 12 \left ( y^2-y^3 \right ) \), and \( x=0 \) about \( y \)-axis.
Solution:
\( V = \int \limits _0 ^1 \pi \left ( 12 \left ( y^2 - y^3 \right ) \right )^2 dy \)
\( =144 \pi \int \limits _0 ^1 \left (y^4 - 2y^5 +y^6 \right )dy \)
\( = 144\pi \left [ \frac{1}{5}y^5 - \frac{1}{3}y^6 + \frac{1}{7}y^7 \right ] _0 ^1 \)
\( = 144 \pi \left ( \frac{1}{5} - \frac{1}{3} + \frac{1}{7} \right ) \)
\( = \frac{48\pi}{35} \)
\( V = \int \limits _0 ^1 \pi \left ( 12 \left ( y^2 - y^3 \right ) \right )^2 dy \)
\( =144 \pi \int \limits _0 ^1 \left (y^4 - 2y^5 +y^6 \right )dy \)
\( = 144\pi \left [ \frac{1}{5}y^5 - \frac{1}{3}y^6 + \frac{1}{7}y^7 \right ] _0 ^1 \)
\( = 144 \pi \left ( \frac{1}{5} - \frac{1}{3} + \frac{1}{7} \right ) \)
\( = \frac{48\pi}{35} \)
