# Area of Surface of Revolution

## Homework Statement

Find area of surface generated by revolving about x-axis.

y=x^3/3 1<=x<=sqrt(7)

find f'(x) = x^2

## The Attempt at a Solution

A = integral[ (x^3/3) * [(1+(x^2)^2] ^(1/2) ] ]dx
= integral[ (x^3/3) * [(1+x^4) ^(1/2) ]dx
I just don't know where to go from here.

Any help is appreciated,
Thank You.

## Answers and Replies

Mark44
Mentor
Ordinary substitution: let u = 1 + x^4.

I'm assuming that you have the correct integrand. If so, this substitution will help you out.

## Homework Statement

Find area of surface generated by revolving about x-axis.

y=x^3/3 1<=x<=sqrt(7)

find f'(x) = x^2

## The Attempt at a Solution

A = integral[ (x^3/3) * [(1+(x^2)^2] ^(1/2) ] ]dx
= integral[ (x^3/3) * [(1+x^4) ^(1/2) ]dx
I just don't know where to go from here.

Any help is appreciated,
Thank You.

You forgot the $$2\pi$$ in your equation, but you probably have that on paper. Here's what the integral should look like.

$$A=\frac{2\pi}{3}\int^{\sqrt{7}}_{1}{x^{3}\sqrt{1+x^{4}}dy$$

You forgot the $$2\pi$$ in your equation, but you probably have that on paper. Here's what the integral should look like.

$$A=\frac{2\pi}{3}\int^{\sqrt{7}}_{1}{x^{3}\sqrt{1+x^{4}}dy$$

Oh whoops, ok well having that in there, how could I continue this problem?

EDIT: I see how this can be solved, substition method.

Thank You.

Last edited:
Do what Mark44 said to do.

$$u=1+x^{4}$$

$$du=4x^{3}dx$$

Last edited:
Mark44
Mentor
Do what Mark44 said to do.

$$u=1+x^{4}$$

$$du=\frac{x^{3}}{4}dx$$
Make that du = 4x3dx

Make that du = 4x3dx

Haha oops.