Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by $y = x ^ { \frac { 3 } { 4 } } , y = 1$, and $x = 0$ about the $y$-axis.
A) $V = \pi \int _ { 0 } ^ { 1 } y ^ { \frac { 8 } { 3 } } d y = \frac { 3 } { 11 } \pi$
B)$V = \pi \int _ { 0 } ^ { 1 } y ^ { \frac { 8 } { 3 } } d y = \frac { 3 } { 22 } \pi$
C)$V = \pi \int _ { 0 } ^ { 1 } y ^ { \frac { 4 } { 3 } } d y = \frac { 3 } { 11 } \pi$
D)$V = \pi \int _ { 0 } ^ { 1 } y ^ { \frac { 3 } { 4 } } d y = \frac { 3 } { 22 } \pi$
E)$V = \pi \int _ { 0 } ^ { 1 } y ^ { \frac { 3 } { 8 } } d y = \frac { 3 } { 11 } \pi$

Question

Quizplus · Accepted Answer

The correct integral setup for the volume of the solid formed by revolving the region around the y-axis is found by solving for $x$ in terms of $y$, which gives $x = y^{\frac{4}{3}}$. The volume is then given by $V = \pi \int_{0}^{1} x^2 dy = \pi \int_{0}^{1} y^{\frac{8}{3}} dy$, which evaluates to $\frac{3}{11}\pi$.

Set Up and Evaluate the Integral That Gives the Volume y=x34,y=1y = x ^ { \frac { 3 } { 4 } } , y = 1y=x43​,y=1

Set Up and Evaluate the Integral That Gives the Volume $y = x ^ { \frac { 3 } { 4 } } , y = 1$