Question

Identify the definite integral that represents the area of the surface formed by revolving the graph of f(x)=x^2 on the interval [0, sqrt(2)] about the x-axis

Answer 1

So, for the area you need:
\[2\pi \int\limits_a^b r(x)\sqrt{1+f'(x)^2}dx\]
So your radius is just going to be the function x^2
Then f'(x)=2x
So you have:
\[2\pi \int\limits_0^\sqrt{2}x^2\sqrt{1+2x}dx\]

Answer 2

wouldn't it be \[2\pi \int\limits_{0}^{\sqrt{2}}x^2\sqrt{1+2x^2}dx\] ?

Answer 3

You're right except it would be (2x)^2=4x^2

Sorry for that D:

Answer 4

I just forgot to square it.

Answer 5

yeah, 4x^2 is what I meant too, haha.