How to Find the Surface Area of a Revolution?

[TheLegace]

Homework Statement

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

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

Homework Equations

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.

 

[Mark44]

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

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

 

[nickmai123]

Homework Statement

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

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

Homework Equations

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.

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

 

[TheLegace]

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.

 

[nickmai123]

Do what Mark44 said to do.

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

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

 

[Mark44]

Do what Mark44 said to do.

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

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

Make that du = 4x³dx

 

[nickmai123]

Make that du = 4x³dx

Haha oops.