# Volume around $y$ axis

To find the volume of the solid of revolution around $$y$$ bounded by $$y=x^2,\quad y=x-2$$ and the lines $$y=0$$ and $$y=1$$, I did as follows: since the region is

Then, the volume is: $$2\pi\cdot\left(\int_{0}^{1}x^3 dx+\int_{1}^{2}xdx+\int_{2}^{3}x(3-x)dx\right)=\frac{35\pi}{6}$$

• Why don’t you see if you get the same with the washer method on $y \in [0,1]$ to verify? Commented May 10 at 2:28
• @RobinSparrow All I want is to know if it's correct, thanks. Commented May 10 at 2:30
• It looks fine. But it would have taken less time than typing the post to do the integration on y via washer to see if you independently got the same answer… Commented May 10 at 2:36

## 1 Answer

Below is an alternative approach, leading to the same result.

You can first calculate the volume of the solid of revolution around $$y$$ bounded by $$y=x^2,\quad y=1,\quad x=0,$$ and then calculate the volume of the frustum of the cone, finally calculate the difference between the two.

The volume of the solid of revolution around $$y$$ bounded by $$y=x^2,\quad y=1,\quad x=0$$ is $$\pi\int_0^1 ydy=\left.\frac\pi2y^2\right|_0^1=\frac\pi2.$$

The volume of the frustum of the cone is $$\frac13\pi\cdot3^2\cdot3-\frac13\pi\cdot2^2\cdot2=\frac{19}3\pi.$$

Finally, the desired volume is $$\frac{19}3\pi-\frac\pi2=\boxed{\frac{35}6\pi}.$$ We get the same answer.