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.