Definite integrals: Calculating Volume Suppose $D$ is the region in the $xy$-plane bounded by the parabola $y=x^2$ and the line $y=2x$. Find the volume of the solid generated by rotating $D$ about 
1)  $x$-axis 
2)  $y$-axis.
Are the answers of 1) and 2) $64π/15$ and  $8π/3$ respectively?
Though it sounds silly, would anyone tell what how to solve this problem?
 A: I will do (1) as an example, from first principles:
Since we are revolving around the x-axis, it will be convenient to describe the volume we are integrating in cylindrical coordinates, so let \begin{equation} y=r \cos{\theta},\; z=r \sin{\theta},\;x=x,\end{equation} then we have the parabola and line described in terms of parameter $r$ with $x$ the dependent variable: \begin{equation} x=\sqrt{r}, \; x=\frac{r}{2}, \end{equation} and within these regions we have $r$ varying from $0$ to where we have $\sqrt{r}=r/2$, solving this equation, we have $r=4$ is the upper limit. And of course $\theta$ varies from $0$ to $2\pi$. So taking all of this together we have the following integration region: \begin{equation} D=\{(r,\theta,x):0\leq \theta \leq 2\pi, \; 0 \leq r \leq 4, \; \frac{r}{2} \leq x \leq \sqrt{r}\}, \end{equation} and since we are using cylindrical coordinates we must include the Jacobian of the coordinate transformation (read more here): in this case $r$, when integrating:
\begin{eqnarray}\text{Vol}(D) &=& \int_0^{2\pi}\int_0^4\int_{r/2}^{\sqrt{r}}r\text{ d}x\text{d}r\text{d}\theta. \end{eqnarray}
So then, calculating the integral, you will get $64\pi/15$, which is indeed the correct answer for number (1). 
A: Hints: 
For a solid generated by rotating $F(x)$ about x axis:
$$V=\int^b_a\pi\left(F(x)\right)^2dx$$
since $y=F(x)$, $x=F^{-1}(y)$, so, for a solid generated by rotating $F(x)$ about y axis:
$$V=\int^b_a\pi\left(F^{-1}(y)\right)^2dy$$
A: Is this a calc II volume of revolution homework problem?


*

*$$\pi\cdot \int_0^2 \left((2x)^2-(x^2)^2\right)\ dx$$

*$$ 2\pi\cdot \int\left(x\cdot(2x-x^2)\right)\ dx$$
