Area of Surface of Revolution

  • Thread starter TheLegace
  • Start date
  • #1
TheLegace
27
0

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.
 

Answers and Replies

  • #2
36,338
8,294
Ordinary substitution: let u = 1 + x^4.

I'm assuming that you have the correct integrand. If so, this substitution will help you out.
 
  • #3
nickmai123
78
0

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 [tex]2\pi[/tex] in your equation, but you probably have that on paper. Here's what the integral should look like.

[tex]A=\frac{2\pi}{3}\int^{\sqrt{7}}_{1}{x^{3}\sqrt{1+x^{4}}dy[/tex]
 
  • #4
TheLegace
27
0
You forgot the [tex]2\pi[/tex] in your equation, but you probably have that on paper. Here's what the integral should look like.

[tex]A=\frac{2\pi}{3}\int^{\sqrt{7}}_{1}{x^{3}\sqrt{1+x^{4}}dy[/tex]

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:
  • #5
nickmai123
78
0
Do what Mark44 said to do.

[tex]u=1+x^{4}[/tex]

[tex]du=4x^{3}dx[/tex]
 
Last edited:
  • #7
nickmai123
78
0

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