Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

ahhh, that makes sense

Answer 7

Thanks!

Answer 8

no problem