Question

The region bounded by y=(x^2)+x-2 and y=0 is rotated about the x axis. Find the volume of the resulting solid by any method.

Answer 1

Find points of intersection.
\[0=(x+2)(x-1) \]
x=-2 or x=1
Find y^2
\[\large y^2=(x+2)^2(x-1)^2\]
\[\large \pi\int\limits_{-2}^{1}y^2dx=\pi\int\limits_{-2}^{1}(x+2)^2(x-1)^2dx\]
Use substitution.
Let u=(x-1)^2
\[u^{\frac{1}{2}}=x-1\]
\[u^{\frac{1}{2}}+1=x\]
\[\frac{1}{2\sqrt{u}}du=dx\]
\[=\large \pi\int\limits_{-2}^{1}(u^{\frac{1}{2}}+3)^2u\frac{1}{2\sqrt{u}}du\]
\[=\large \pi\int\limits_{-2}^{1}(u^{\frac{1}{2}}+3)^2\frac{\sqrt{u}}{2}du\]
\[=\large \frac{\pi}{2}\int\limits_{-2}^{1}u^{\frac{1}{2}}(u^{\frac{1}{2}}+3)^2du\]
\[=\large \frac{\pi}{2}\int\limits_{-2}^{1}u^{\frac{1}{2}}(u+6u^{\frac{1}{2}}+9)du\]
\[=\large \frac{\pi}{2}\int\limits_{-2}^{1}u^{\frac{3}{2}}+6u+9u^{\frac{1}{2}}du\]
\[=\large \frac{\pi}{2}[\frac{2}{5}u^{\frac{5}{2}}+3u^2+6u^{\frac{3}{2}}]\]

Answer 2

Just squaring the quadratic may have been more straightforward here.

Answer 3

\[\large=\frac{\pi}{2}[\frac{2}{5}(x-1)^5+3(x-1)^4+6(x-1)^3]\]
\[\large=\frac{\pi}{2}[0-(-\frac{486}{5}+243-162)]\]
\[\large=\frac{\pi}{2}[\frac{81}{5}]\]
\[\large=\frac{81\pi}{10}u^3\]