MHB Double Integral Problem/ Surface Area of parametric surface

Click For Summary
The discussion focuses on solving a double integral related to the surface area of a parametric surface. The user has successfully computed the cross product but is struggling with the double integral. A response clarifies that the inner integral can be evaluated first, simplifying the process. The final expression for the integral is presented as π times the integral of the square root of (1 + u²) from 0 to 1. This highlights a method for tackling double integrals in surface area calculations.
Dizzy1
Messages
4
Reaction score
0
Hi I'm doing a surface area problem with a parametric surface and I got the cross product but I can't figure out the double integral.

I found the solution online but with no explanation, so can someone explain how to solve this integral:

View attachment 2449

thank you!
 

Attachments

  • Screen Shot 2014-05-06 at 2.24.41 PM.png
    Screen Shot 2014-05-06 at 2.24.41 PM.png
    2.8 KB · Views: 97
Physics news on Phys.org
Dizzy said:
Hi I'm doing a surface area problem with a parametric surface and I got the cross product but I can't figure out the double integral.

I found the solution online but with no explanation, so can someone explain how to solve this integral:

View attachment 2449

thank you!

Because the inner integral is just a number you can immediately do the outer integral so:

$$\int_0^{\pi}\ \int_0^1\sqrt{1+u^2}du\ dv=\pi \int_0^1\sqrt{1+u^2}du$$

.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K