# Double Integration in Polar Coordinates

• maxhersch
In summary, the integral is converted to polar coordinates by changing the bounds of integration to r from 0 to 6 and theta from 0 to pi/2. The integrand is simplified to just theta, but a factor of r is missing in the final integral. The correct result is 3pi^2/4.
maxhersch

## Homework Statement

Integrate by changing to polar coordinates:

## \int_{0}^6 \int_{0}^\sqrt{36-x^2} tan^{-1} \left( \frac y x \right) \, dy \, dx ##

## Homework Equations

## x = r \cos \left( \theta \right) ##
## y = r \sin \left( \theta \right) ##

## The Attempt at a Solution

So this is a quarter of a circle in the first quadrant of an xy-coordinate system with a radius of 6. So the bounds for the integral in polar coordinates will be r from 0 to 6 and ## \theta ## from 0 to ## \frac{ \pi }{2} ## .

## tan^{-1} \left( \frac y x \right) ## is just ## \theta ## so the integral becomes:

## \int_{0}^6 \int_{0}^ \frac{ \pi }{2} \theta \, d \theta \, dr ##

I worked this out and got ## \frac{3 \pi ^2}{4} ## . I checked it on Wolfram Alpha and got the same result. I plugged the original integral into Wolfram Alpha and got ## \frac{9 \pi ^2}{4} ## so somehow I lost a factor of 3 somewhere. Clearly I integrated properly so I must have made a mistake converting to polar coordinates.

Any help would be appreciated. Thanks.

In your conversion to polar the elemental area in Cartesian is dA = dy*dx

However in polar it should be r*dr*dθ

You have forgot a factor of ##r##. The area element ##dxdy## should be replaced by ##rdrd\theta##, which is obtained from the Jacobian for the transformation to polar coordinates. I hope this helps.

rock.freak667 said:
In your conversion to polar the elemental area in Cartesian is dA = dy*dx

However in polar it should be r*dr*dθ

Yep, just remembered that right as your replied. Thanks

## 1. What is double integration in polar coordinates?

Double integration in polar coordinates is a mathematical method used to calculate the area under a curve or surface in a polar coordinate system. It involves integrating a function over a region in the polar plane, and then integrating the result over the entire region.

## 2. How is double integration in polar coordinates different from Cartesian coordinates?

In Cartesian coordinates, the x and y coordinates represent distances along the x and y axes, respectively. In polar coordinates, the r and θ coordinates represent a distance from the origin and an angle from the positive x-axis, respectively. This means that the integration limits and the equations used for integration are different in polar coordinates compared to Cartesian coordinates.

## 3. What is the formula for double integration in polar coordinates?

The formula for double integration in polar coordinates is ∬f(r,θ) rdrdθ, where f(r,θ) is the function being integrated and the limits of integration for r and θ depend on the region being integrated over.

## 4. What are the applications of double integration in polar coordinates?

Double integration in polar coordinates is commonly used in physics and engineering to calculate the mass, center of mass, and moment of inertia of objects with circular or symmetric shapes. It is also used in calculus to solve problems involving polar coordinates, such as finding the area of a polar curve or the volume of a polar solid.

## 5. What are some tips for solving double integration problems in polar coordinates?

Some tips for solving double integration problems in polar coordinates include carefully setting up the limits of integration, converting the function to polar coordinates, and using trigonometric identities to simplify the integrand. It is also important to visualize the region being integrated over and to understand the relationship between polar and Cartesian coordinates.

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