# Double Surface Integrals in Polar Coordinates

• allllllll
In summary, the conversation discusses finding the surface area of a cone above a region in the xy-plane with a given area. After some simplification and conversion to polar coordinates, the integrand is found to be a constant, resulting in a final answer of the constant multiplied by the area of the given region. There is also a brief discussion about a potential error in the original equation.
allllllll

## Homework Statement

Find the surface area of the cone z=3x^2+y^2and above a region in the xy-plane with area 4.

## Homework Equations

double integral sqrt( (dz/dx)^2 + (dz/dy)^2 +1)

## The Attempt at a Solution

I was able to simplify the equation, I just don't know what to do with the area part of the question:

dbl int sqrt ( (3x/sqrt(x^2+y^2))^2 + (3y/sqrt(x^2+y^2))^2 + 1)
dbl int sqrt (9 (x^2+y^2)/(x^2+y^2) + 1 )
dbl int sqrt (10)

change into polar coordinates

integral [0-1] dtheta integral [0-2pi] sqrt(10) r dr
= 1*2pi sqrt(10)

what am I doing wrong?

That's not a cone It's a paraboloid. Did you actually mean to say z^2=3*(x^2+y^2)? That's a cone. I'm just guessing that from your work. But if you do that you don't get sqrt(10) for the integrand of your double integral. And you can't integrate over the xy domain using polar coordinates. You don't even know what the xy domain IS! All you know is it has area 4. Is that enough?

yeah sorry should have said z=3sqrt(x^2+y^2)..

answer is sqrt(10)*4.. don't know why

You have sqrt(10) for the integrand of the surface integral. It's a constant. If you integrate a constant over a domain of area 4, what do you get? I can repeat this if it's too simple. I really don't mind. I can cut and paste.

allllllll said:
yeah sorry should have said z=3sqrt(x^2+y^2)..
So $\sqrt{x^2+ y^2}= z/3$

and so (3x/sqrt(x^2+y^2))^2 + (3y/sqrt(x^2+y^2))^2 + 1)= (9x+ 9y+ z)/z.

HallsofIvy said:
So $\sqrt{x^2+ y^2}= z/3$

and so (3x/sqrt(x^2+y^2))^2 + (3y/sqrt(x^2+y^2))^2 + 1)= (9x+ 9y+ z)/z.

Isn't that 9x^2/(x^2+y^2)+9y^2/(x^2+y^2)+1=9*(x^2+y^2)/(x^2+y^2)+1=9+1=10?

Yes, It is. We clearly have solved this problem! I hope allllllll has too!

alllllllllllllllll had the sqrt(10) from the beginning. Just couldn't figure out how to handle the domain. Yes, let's hope it's got.

## 1. What are double surface integrals in polar coordinates?

Double surface integrals in polar coordinates are a type of mathematical calculation used in multivariable calculus. They involve integrating over a two-dimensional region in polar coordinates, where the variables are represented by radial distance and angle from the origin. This type of integration is commonly used in physics and engineering to calculate the area or volume of curved surfaces.

## 2. How do you convert a double surface integral from rectangular to polar coordinates?

To convert a double surface integral from rectangular to polar coordinates, you can use the following formula:
R f(x,y) dA = ∬D f(r,θ) r dr dθ
where D is the region in polar coordinates and r and θ are the variables. The limits of integration for r and θ can be determined by graphing the region and finding the points where the curves intersect.

## 3. What is the difference between a single and double surface integral in polar coordinates?

A single surface integral in polar coordinates involves integrating over a one-dimensional curve, while a double surface integral involves integrating over a two-dimensional region. In polar coordinates, the variables are represented by radial distance and angle from the origin, whereas in rectangular coordinates, they are represented by x and y coordinates.

## 4. What are some real-world applications of double surface integrals in polar coordinates?

Double surface integrals in polar coordinates are commonly used in physics and engineering to calculate the mass, center of mass, and moment of inertia of objects with curved surfaces. They are also used in fluid mechanics to calculate the flow rate of a fluid through a curved surface.

## 5. What are some common mistakes to avoid when working with double surface integrals in polar coordinates?

Some common mistakes to avoid when working with double surface integrals in polar coordinates include forgetting to convert the function being integrated to polar coordinates, using the wrong limits of integration, and forgetting to include the extra r factor in the integrand. It is also important to carefully graph the region before setting up the integral to ensure the correct limits of integration are used.

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