Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the $x$-axis.
$y=x^{3/2}$, $y=8$, $x=0$
What I did:
$y=x^{3/2}\Rightarrow x=y^{2/3}$
$Circumference= 2\pi(8-y)$
$Height=(4-y^{2/3})$
$Surface Area =2\pi(8-y)(4-y^{2/3})$
$Volume=2\pi\int _{ 0 }^{ 8 }{ (8-y)(4-y^{2/3})dy} $
$=2\pi\int _{ 0 }^{ 8 }{ 32-8y^{2/3}-4y+y^{5/3}dy}$
$=2\pi[32y-\frac{24y^{5/3}}{5}-2y^2+\frac{3y^{8/3}}{8}]_{0}^{8}$
$=2\pi[32(8)-\frac{24(8)^{5/3}}{5}-2(8)^2+\frac{3(8)^{8/3}}{8}]-2\pi[0]$
$=2\pi[256-\frac{768}{5}-128+\frac{768}{8}]$
$=2\pi(\frac{352}{5})$
$=\frac{704\pi}{5}$
My final answer seems to be wrong, but I am not sure where I went wrong in my (wrong) solution for this problem. Any help/guidance would be greatly appreciated.
Best Answer
Since the revolution is around the $x$-axis, each cylindrical shell at $x$ for $x\in [0,4]$ (where $4=8^{3/2}$) is actually an annulus with inner radius $r(x)=x^{3/2}$ and outer radius $R(x)=8$.
The volume of this shell is $(\pi R^2(x) -\pi r^2(x))\cdot dx$. Hence the total volume can be obtained by "summing" the volumes of all these thin shells, $$\int_{x=0}^4 (\pi R^2(x) -\pi r^2(x))\cdot dx=\pi\int_{0}^4 (8^2 -x^3)\cdot dx=\pi \left[64x-\frac{x^4}{4}\right]_0^4= 192\pi.$$
Note that if the revolution is around the $y$-axis, the solid is different and its volume is $$\int_{y=0}^8 \pi (y^{2/3})^2\cdot dy=\pi \left[\frac{y^{7/3}}{7/3}\right]_0^8=\frac{384\pi}{7}.$$