0
$\begingroup$

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 enter image description here

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

$\endgroup$
3
  • $\begingroup$ Why don’t you see if you get the same with the washer method on $y \in [0,1]$ to verify? $\endgroup$ Commented May 10 at 2:28
  • $\begingroup$ @RobinSparrow All I want is to know if it's correct, thanks. $\endgroup$
    – mvfs314
    Commented May 10 at 2:30
  • 2
    $\begingroup$ 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… $\endgroup$ Commented May 10 at 2:36

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .