Surface Area [Double Integral]

  • Thread starter 1d20
  • Start date
  • #1
12
0
I’m doing a lot of double integrals to find surface area problems, and I don’t think I’m setting them up quite right. For example,

“Find the surface area of the portion of the sphere [itex] x^2 + y^2 + z^2 = 25 [/itex] inside the cylinder [itex]x^2 + y^2 = 9[/itex].”

I converted the sphere to a function of z: [itex]\sqrt{25 - x^2 - y^2} [/itex]

Then I found the partial derivatives:
fx = [itex]\frac{-x} {\sqrt{25 - x^2 - y^2}} [/itex]
fy = [itex]\frac{-y} {\sqrt{25 - x^2 - y^2}} [/itex]

Then I set up the integral:

SA = [itex]\int \int \sqrt{1 + (fx)^2 + (fy)^2} dx dy [/itex]

SA = [itex]\int \int \sqrt{1 + \frac{x^2} {25 - x^2 - y^2} + \frac{y^2} {25 - x^2 - y^2}} dx dy [/itex]

Adding fractions, I get:

SA = [itex]\int \int \sqrt{ \frac{25} {25 - x^2 - y^2} } dx dy [/itex]

Converting to polar, I get:

SA = [itex]\int_0^3 \int_0^{2\pi} r\sqrt{ \frac{25} {25 - r^2} } d\theta dr [/itex]

(That pi should be a part of the integral range, but I can’t find the fraking index of forum code.)

I can't think of a way to integrate this, so I think I didn't set it up right.
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
Looks ok to me. You've got an extra r^2 in the numerator of the last expression, but I'll assume that's a typo since the expression before it looks good. So you want to integrate r/sqrt(25-r^2). Just use a u-substitution.
 
  • #3
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,557
767
I’m doing a lot of double integrals to find surface area problems, and I don’t think I’m setting them up quite right. For example,

“Find the surface area of the portion of the sphere [itex] x^2 + y^2 + z^2 = 25 [/itex] inside the cylinder [itex]x^2 + y^2 = 9[/itex].”

I converted the sphere to a function of z: [itex]\sqrt{25 - x^2 - y^2} [/itex]
You mean a function of x and y and you would write [itex]z = f(x,y)=\sqrt{25 - x^2 - y^2} [/itex], which gives only the top half of the sphere.
Then I found the partial derivatives:
fx = [itex]\frac{-x} {\sqrt{25 - x^2 - y^2}} [/itex]
fy = [itex]\frac{-y} {\sqrt{25 - x^2 - y^2}} [/itex]

Then I set up the integral:

SA = [itex]\int \int \sqrt{1 + (fx)^2 + (fy)^2} dx dy [/itex]

SA = [itex]\int \int \sqrt{1 + \frac{x^2} {25 - x^2 - y^2} + \frac{y^2} {25 - x^2 - y^2}} dx dy [/itex]

Adding fractions, I get:

SA = [itex]\int \int \sqrt{ \frac{25} {25 - x^2 - y^2} } dx dy [/itex]

Converting to polar, I get:

SA = [itex]\int_0^3 \int_0^2pi r\sqrt{ \frac{r^2} {25 - r^2} } dO dr [/itex]

(That pi should be a part of the integral range, but I can’t find the fraking index of forum code.)
.
OK until that last step. Why did you replace the 25 by r2? And when you want more than a single letter in the limit use {} to enclose, like this:

[tex]\int_0^3 \int_0^{2\pi} r\sqrt{ \frac{25} {25 - r^2} } d\theta dr [/tex]

You can right click on that formula to see how I changed it in a couple of spots. Then don't forget, this is just the top half.

[Edit] Dang!! Dick beat me to it again.
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
[Edit] Dang!! Dick beat me to it again.
It's good you came in. I missed that extra factor of 2.
 
  • #5
12
0
Thanks! It was r being in the denominator inside a radical that threw me off.

While I'm here, does this forum have a list of all the symbol codes? I'd post here more often if I didn't have to 'cheat' from other threads every time I want to post.
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,557
767
Thanks! It was r being in the denominator inside a radical that threw me off.

While I'm here, does this forum have a list of all the symbol codes? I'd post here more often if I didn't have to 'cheat' from other threads every time I want to post.
You can read my signature file for one way to get some of them. Or if you are using TeX, you can click on the [itex]\Sigma[/itex] icon above in the advanced edit box for a TeX editor.

And I just now noticed there are symbols to the right of the advanced edit box.
 

Related Threads on Surface Area [Double Integral]

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
10
Views
1K
Replies
9
Views
3K
Replies
3
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
2
Views
8K
  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
575
  • Last Post
Replies
2
Views
5K
Top