Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^5, y=0, x=1, x=8; about the y-axis.

Answer 1

if you wanted to rotate f(x) around the y axis from x=a to x=b then the volume would be
$\pi \int\limits^a_b {(f(x))^2)} \, dx$


that would be

$\pi \int\limits^8_1 {(\frac{1}{x^5})^2} \, dx$
$\pi \int\limits^8_1 {\frac{1}{x^{10}}} \, dx$
1/x^10=x^-10
integrate
-1/(9x^9)
so
$\pi \int\limits^8_1 {\frac{1}{x^{10}}} \, dx$=
$\pi [\frac{-1}{9x^{9}}]^8_1=\pi (\frac{-1}{9(8^{9})}-\frac{-1}{9(1^{9})})=$
$\pi (\frac{-1}{9(8^{9})}+\frac{1}{9(1^{9})})=\pi (\frac{-1}{9(8^{9})}+\frac{1}{9})=$
$\pi (\frac{-1}{9(8^{9})}+\frac{8^9}{9(8^9)})=\pi (\frac{8^9-1}{9(8^{9})})=$
$(\frac{\pi8^9-\pi}{9(8^{9})})=\frac{134217727 \pi}{1207959552}$
that's the volume of the solid