Surface Area [Double Integral]

[1d20]

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 x^2 + y^2 + z^2 = 25 inside the cylinder x^2 + y^2 = 9.”

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

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

Then I set up the integral:

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

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

Adding fractions, I get:

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

Converting to polar, I get:

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

(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.

 

[Dick]

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.

 

[LCKurtz]

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 x^2 + y^2 + z^2 = 25 inside the cylinder x^2 + y^2 = 9.”

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

You mean a function of x and y and you would write z = f(x,y)=\sqrt{25 - x^2 - y^2}, which gives only the top half of the sphere.

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

Then I set up the integral:

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

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

Adding fractions, I get:

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

Converting to polar, I get:

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

(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:

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

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.

 

[Dick]

[Edit] Dang! Dick beat me to it again.

It's good you came in. I missed that extra factor of 2.

 

[1d20]

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.

 

[LCKurtz]

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 \Sigma 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.