<s>[INST] Volume of Solid Revolution region bounded by $$y=x$$ and $$y=x^2$$
a) find the volume of the solid of revolution formed by revolving R region about the line $x=2$.
b) find the volume of the solid of revolution formed by revolving R region about the line $y=b$ where $b > 1$ is a constant
c) Find any values of a for which the volume of solid in part b is same as volume of solid in part a. Approximate to two decimal places and may use a calculator
I got $\frac \pi2$  for my answer for a but I feel like thats wrong. I integrated with respect to y and found the bounds as 0 to 1 but after calculating I get $\frac \pi2$ for an answer. Please explain in detail how to do these parts and let me know what I did wrong in part a (if its wrong).
 [/INST] we have that the region is bounded by $y=x$ and $y=x^2$, finding the intersections
$\left.\begin{matrix}y=x\\y=x^2\end{matrix}\right\} x=x^2\iff x=0\vee x=1$
so, them intercepts on $(0,0)$ and $(1,1)$, then:
a) then the volume of revolution about the line $x=2$, we have that $x=y,x=\sqrt{y}$, then.
for $0\le y\le 1\Rightarrow 2\ge\sqrt{y}\ge y$, wich implies that $(y-2)^2\ge(\sqrt{x}-2)^2$, then we gets that
$$\begin{align}V&=\pi\int_{0}^{1}\left|(y-2)^2-(\sqrt{y}-2)^2\right|dy\\
&=\pi\int_{0}^{1}(y-2)^2-(\sqrt{y}-2)^2dy\\
&=\pi\int_{0}^{1}(y^2-4y+4)-(y-4\sqrt{y}+4)dy\\
&=\pi\int_{0}^{1}y^2-4y+4-y+4\sqrt{y}-4dy\\
&=\pi\int_{0}^{1}y^2-5y+4\sqrt{y}dy\\
&=\pi\left[\frac{y^3}{3}-\frac{5y^2}{2}+\frac{8y\sqrt{y}}{3}\right]_{0}^{1}\\
&=\pi\left(\frac{1}{3}-\frac{5}{2}+\frac{8}{3}\right)\\
&=\pi\frac{2-15+16}{6}\\
&=\frac{\pi}{2}\end{align}$$
b) for this, the solid is revolving over line $y=a$, with $a>1$, then we got:
for $0\le x\le1\Rightarrow a\ge x\ge x^2$, then $(x^2-a)^2\ge(x-a)^2$, then we gets that
$$\begin{align}V&=\pi\int_{0}^{1}\left|(x-a)^2-(x^2-a)\right|dx\\
&=\pi\int-\left[(x-a)^2-(x^2-a)\right]dx\\
&=-\pi\int(x^2-2ax+a^2)-(x^4-2ax^2+a^2)dx\\
&=-\pi\int x^2-2ax+a^2-x^4+2ax^2-a^2dx\\
&=-\pi\int-x^4+(1+2a)x^2-2axdx\\
&=-\pi\left[-\frac{x^5}{5}+\frac{(1+2a)x^3}{3}-ax^2\right]_{0}^{1}\\
&=-\pi\left(-\frac{1}{5}+\frac{1+2a}{3}-a\right)\\
&=-\pi\frac{-3+5(1+2a)-15a}{15}\\
&=-\pi\frac{-3+5+10a-15a}{15}\\
&=-\pi\frac{2-5a}{15}\\
&=\pi\frac{5a-2}{15}
\end{align}$$
c) for $c$ we have
$$\frac{\pi}{2}=\pi\frac{5a-2}{15}\\
15=2(5a-2)\\
15=10a-4\\
10a=15+4=19\\
a=\frac{19}{10}=1.9
$$
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