Double Integral - Going from Cartesian to Polar

In summary, the conversation is about converting a Cartesian integral into an equivalent polar integral and then evaluating it. The person is having trouble with the limits and has posted two examples in the attachment. The other person suggests sketching a graph of the region in the x-y plane to help with finding the theta limits. They also mention that the teacher did not explain how to convert to cylindrical and spherical coordinates, causing confusion for the person. They eventually come to the conclusion that the upper limit for the second example should be 2*sec(theta) and the theta limits may not be correct.
  • #1
erok81
464
0

Homework Statement



See attachment.

Change the Cartesian integral into an equivalent polar integral, then evaluate the integral.

I have no problems at all converting the actual function I am integrating or the integration itself, it is just the limits I cannot do.

I've posted two examples in my attachment. The first one I have the answer for and I can get three of them but the 6csc theta I cannot figure how to get that. The other I've posted the three I can do (but don't know if it is right because I have no answer for this one) but still that fourth one I can't get get.

I think the problem is our teacher gave us an assignment to convert Cartesian to cylindrical to spherical but didn't explain how to do it or give us any answers. You can imagine that was a total waste of time - "here do this assignment but I am not going to tell you how or give you the answers." :eek:
 

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  • #2
Why don't you sketch a graph of the region in the x-y plane? Then the theta limits of pi/4 to pi/2 shouldn't be any surprise. Now you want to find the r limits as a function of theta. y=r*sin(theta). The limits for y are 0 and 6. Solve for r. An upper limit of 6*csc(theta) shouldn't be a surprise either.
 
  • #3
That makes sense.

I guess the part I am confused about is the cartesian coordinates are integrated as dx dy so I though you'd covert dx to dr and dy to d(theta) since they are in that order.

I thought I got it but after I read it I took my dog out and came back to work on it and now it doesn't make sense.
 
  • #4
Or is it just in this case you know that y=rsin(theta) so you can pull the y values from the other integral limits?
 
  • #5
Oh...maybe I do get it.

In my second example is the upper dr limit 2sec(theta)?

Now that I think about it more, I remember the teacher said carve with r and sweep with theta. If I do that my post #4 is irrelevant and that part makes sense.
 
  • #6
erok81 said:
Oh...maybe I do get it.

In my second example is the upper dr limit 2sec(theta)?

Now that I think about it more, I remember the teacher said carve with r and sweep with theta. If I do that my post #4 is irrelevant and that part makes sense.

Yes, to 2*sec(theta). But in the second example I don't think the theta limits are right. Did you draw the region?
 

1. What is a double integral in Cartesian coordinates?

A double integral is a mathematical concept that involves finding the area under a surface defined by two variables (x and y) in a Cartesian coordinate system. It is represented by the symbol ∫∫f(x,y) dxdy.

2. What is the purpose of converting a double integral from Cartesian to Polar coordinates?

Converting a double integral from Cartesian to Polar coordinates can make it easier to evaluate the integral, especially when the region of integration has a circular or polar symmetry. It also allows for a more intuitive understanding of the integral as it relates to polar coordinates.

3. What is the formula for converting a double integral from Cartesian to Polar coordinates?

The formula for converting a double integral from Cartesian to Polar coordinates is ∫∫f(x,y) dxdy = ∫∫f(r,θ) rdrdθ, where r represents the distance from the origin and θ represents the angle formed with the positive x-axis.

4. Can any double integral be converted from Cartesian to Polar coordinates?

No, not all double integrals can be converted from Cartesian to Polar coordinates. The integral must have a region of integration that has a circular or polar symmetry in order for the conversion to be valid.

5. What are some real-world applications of double integrals in Cartesian and Polar coordinates?

Double integrals in Cartesian coordinates are commonly used in physics and engineering to calculate the mass, center of mass, and moments of inertia of a 2D object. Double integrals in Polar coordinates are often used in physics and engineering to calculate the electric field and gravitational potential of a 2D object with circular symmetry.

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