## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 5 years ago I have to find the Surface Area of the solid that is created by the curve: y=1-x^(2) rotated around the y-axis. on the interval 0 less than or equal to x less than or equal to 1

• This Question is Closed
1. anonymous

so far I have $2\pi \int\limits_{0}^{1}$

2. anonymous

$\int\limits_{a}^{b} 2pif(x)*\sqrt(1 + (dy/dx)^2)$

3. anonymous

I brought the 2pi out because it's a constant

4. anonymous

actually, let me correct that, the f(x) will just be x because your x value will be the radius of the solid

5. anonymous

so you should get: $2\pi \int\limits_{0}^{1} x \sqrt{1 + 4x^2}dx$

6. anonymous

ahhh, that makes sense

7. anonymous

Thanks!

8. anonymous

no problem

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy