# Surface Area [Double Integral]

• 1d20
They don't seem to be the same as the ones you get from hitting the Sigma icon though.In summary, the conversation discusses setting up a double integral to find the surface area of a portion of a sphere inside a cylinder. The sphere is converted to a function of z and partial derivatives are found. The integral is then set up and converted to polar form. However, there is a mistake in the last step where the 25 is replaced by r^2. The correct integral is given as \int_0^3 \int_0^{2\pi} r\sqrt{ \frac{25} {25 - r^2} } d\theta dr. The conversation also briefly touches on the available symbol codes for posting on

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

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.

Last edited:
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.

1d20 said:
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 = $\int \int \sqrt{1 + \frac{x^2} {25 - x^2 - y^2} + \frac{y^2} {25 - x^2 - y^2}} dx dy$

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.

 Dang! Dick beat me to it again.

LCKurtz said:
 Dang! Dick beat me to it again.

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

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.

1d20 said:
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.

## 1. What is the definition of surface area in a double integral?

The surface area in a double integral refers to the measure of the total area of a three-dimensional object. It is calculated by integrating the function that defines the surface over a specified region.

## 2. How is surface area represented in a double integral?

Surface area is represented as a double integral in the form ∬f(x,y) dA, where f(x,y) is the function that defines the surface and dA is the differential area element.

## 3. What is the difference between single and double integration in relation to surface area?

Single integration is used to find the area under a curve in two dimensions, while double integration is used to find the volume or surface area of a three-dimensional object. In the case of surface area, double integration is used to find the surface area of a three-dimensional object by integrating over a specified region.

## 4. How is the double integral for surface area evaluated?

The double integral for surface area is evaluated by first setting up the limits of integration for the two variables, which define the region of the surface. Then, the function that defines the surface is integrated with respect to both variables, and the resulting expression is evaluated to find the total surface area.

## 5. What are some real-life applications of double integration for surface area?

Double integration for surface area has many practical applications in fields such as engineering, physics, and computer graphics. It is used to calculate the surface area of objects such as buildings, bridges, and curved surfaces in 3D modeling. It is also used in fluid mechanics to calculate the surface area of irregularly shaped objects such as ships and aircraft wings.