How to Find the Limits of Integration for a Double Integral?

In summary, the conversation is about evaluating a double integral in the first quadrant bounded by the circle x^2+y^2=9. The main focus is on finding the limits of integration, with the expert providing guidance and clarifying that the region is not a square. The final solution is to integrate from 0 to 3 for both x and y, in the order of dy dx.
  • #1
Math10
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Homework Statement


Evaluate the double integral (x+2y)dA, where R is the region in the first quadrant bounded by the circle x^2+y^2=9.

Homework Equations


None.

The Attempt at a Solution


I know how to evaluate the double integral but I just don't know how to find the limits of integration. I know that the radius of the circle is 3.
 
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  • #2
First, what are the limits for x?
 
  • #3
I have no idea. Is it 0 to 3?
 
  • #4
It looks like you do have an idea. That's right.

For each x, what are the limits of y? You may need to draw the area.
 
  • #5
From 0 to 3?
 
  • #6
Math10 said:

Homework Statement


Evaluate the double integral (x+2y)dA, where R is the region in the first quadrant bounded by the circle x^2+y^2=9.

Homework Equations


None.

The Attempt at a Solution


I know how to evaluate the double integral but I just don't know how to find the limits of integration. I know that the radius of the circle is 3.
How would you describe the region R as a set? IOW, R = {(x, y) | <inequality involving x> and <inequality involving y>}.

If you are tempted to write the inequalities as ##0 \le x \le 3, \text{and } 0 \le y \le 3##, resist that temptation! That region would be a square, which the one in this problem is not.

As I said in another of your threads, many times carrying out the integration is the easy part, requiring very little thinking. Figuring out the limits of integration is often the more difficult task. Your textbook should have a number of examples. Have you looked at them?
 
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  • #7
Math10 said:
From 0 to 3?
That would be a square in the first quadrant.
 
  • #8
So y=sqrt(9-x^2) and y= - sqrt(9-x^2)?
 
  • #9
Math10 said:
So y=sqrt(9-x^2) and y= - sqrt(9-x^2)?
Why would y be negative?
 
  • #10
So y=0 to sqrt(9-x^2)?
 
  • #11
Math10 said:
So y=0 to sqrt(9-x^2)?
Yes. Now what are your limits for x?
 
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  • #12
0 to 3?
 
  • #13
Yes. Now you need to figure the order in which you're going to integrate.
 
  • #14
dy dx?
 
  • #16
dy dx.
 
  • #17
OK. Now you're ready to set up the iterated integral and evaluate it...
 
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  • #18
Thank you!
 
  • #19
You're welcome!
 

1. What is a double integral?

A double integral is a mathematical concept used to find the volume under a surface or between two surfaces in three-dimensional space.

2. How do you evaluate a double integral?

To evaluate a double integral, you first need to set up the limits of integration for both the inner and outer integrals. Then, you can use various integration techniques such as substitution, u-substitution, or integration by parts to solve the integral.

3. What is the difference between a single integral and a double integral?

A single integral is used to find the area under a curve in one dimension, while a double integral is used to find the volume under a surface or between two surfaces in two dimensions.

4. Can a double integral be evaluated using a calculator?

Yes, certain calculators have the capability to evaluate double integrals using numerical methods such as Simpson's rule or the trapezoidal rule. However, it is important to note that these methods may not give exact solutions and may be subject to rounding errors.

5. What are some applications of double integrals?

Double integrals have various applications in physics, engineering, and economics. They are used to calculate the work done by a force, find the center of mass of an object, and determine the total profit of a business, among other things.

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