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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.

“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.

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