# Volume of a Solid of Rotation Bounded by Two Curves

I'm having a bit of a hard time with this problem from a Calc 1 Textbook:

Find the volume of the solid obtained by rotating the region bounded by the given curves along the given axis:

$$y=x^2$$

$$x=y^2$$

about the x-axis

My Work:

$$\pi\displaystyle\int_0^1(\sqrt{x}-x^2)^2dx$$

$$\pi\displaystyle\int_0^1(x-2\sqrt{x}x^2+x^4)dx$$

$$\pi(\frac{1}{2}-\frac{4}{7}+\frac{1}{5})$$

$$\frac{9\pi}{70}$$

Book's Answer: $$\frac{3\pi}{10}$$

What did I do wrong? Thanks.

## 1 Answer

What you need to do is first find the volume when $$\sqrt x$$ is rotated, and then subtract the part when $$x^2$$ is rotated, i.e., $$V = \pi\int_0^1 (\sqrt x)^2\, dx - \pi\int_0^1 (x^2)^2\, dx = \frac{3\pi}{10}.$$

What you did is rotate $$\sqrt x - x$$, which will give a different volume.

(Notice that unlike usual integration, this operation of rotating is not linear; i.e. $$\pi\int_a^b (f+g)^2\,dx \neq \pi\int_a^b f^2\,dx + \pi\int_a^b g^2\,dx$$.)

• Didn't I subtract $\sqrt{x}$ and $x^2$?. The book said $\frac{3\pi}{10}$ – N. Bar Jun 22 '19 at 20:41
• Notice what I said about linearity, you cannot subtract them inside the integral because the integrand is squared. That's like assuming $(f+g)^2 = f^2+g^2$. – Luke Collins Jun 22 '19 at 20:42
• Ah, I see. Also, didn't see the edit where you changed the answer. – N. Bar Jun 22 '19 at 20:43
• Yes, I copied the answer wrong from Mathematica. – Luke Collins Jun 22 '19 at 20:44
• Thanks, this really helped. – N. Bar Jun 22 '19 at 20:52