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.