Problem with a double integral

In summary, the conversation discusses how to compute the integral (3x dx dy) by converting it into polar coordinates. The region of integration is divided into two regions, but it is suggested to use only one region considering the polar coordinate system. The region is described by the disk equation r=2cos(θ) and the limits for theta are 0 to (pi/4). It is confirmed that the approach works and the final answer is (3pi/4)+1.
  • #1
Amaelle
310
54
Homework Statement
Compute the integral (3x dx dy), where the region of equation is : (x − 1)^2 + y^2 ≤ 1, 0 ≤ y ≤ x
Relevant Equations
0<=r<=2cos(θ),
I already have the solution in which the region of integration has been divided into two regions
1597153537239.png

but I was wondering if I can only use one region considering the polar coordinate system) the disk equation for me is r=2cos(θ) and the theta goes from 0 to (pi/4)

0<r<2cos(θ) and the 0 <θ<pi/4
the total integral becomes
D3%20r%5E%7B2%7B%7D%7Dcos%5CTheta%20dr%20d%5CTheta.gif


is my approach is correct?

Thanks a lot in advance!
 
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  • #2
Amaelle said:
Homework Statement:: Compute the integral (3x dx dy), where the region of equation is : (x − 1)^2 + y^2 ≤ 1, 0 ≤ y ≤ x
Relevant Equations:: 0<=r<=2cos(θ),

I already have the solution in which the region of integration has been divided into two regions View attachment 267622
but I was wondering if I can only use one region considering the polar coordinate system) the disk equation for me is r=2cos(θ) and the theta goes from 0 to (pi/4)

0<r<2cos(θ) and the 0 <θ<pi/4
the total integral becomes
View attachment 267623

is my approach is correct?

Thanks a lot in advance!
I don't see why it wouldn't work. See if you get the same answer with both approaches. If you do, that would confirm that your approach is correct, and would also be good practice for converting integrals from Cartesian (rectangular) form to polar form.
 
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  • #3
Thanks a lot for your prompt answer, the problem is that I couldn't find the same
Mark44 said:
I don't see why it wouldn't work. See if you get the same answer with both approaches. If you do, that would confirm that your approach is correct, and would also be good practice for converting integrals from Cartesian (rectangular) form to polar form.
Thanks a lot , i did it and it works!
 
  • #4
etotheipi said:
You can do it even without a double integral. You just need to evaluate ##\frac{1}{2}\int (2\cos{\theta})^2 d\theta##$$A = \frac{1}{2} \int_0^{\frac{\pi}{4}} 4\cos^2{\theta} \, d\theta = 2\int_0^{\frac{\pi}{4}} \frac{1}{2} + \frac{1}{2}\cos{2\theta} \, d\theta = \left[ \theta +\frac{1}{2}\sin{2\theta} \right]_0^{\frac{\pi}{4}} = \frac{\pi}{4} + \frac{1}{2}$$
thanks a lot but i think you did a miscalculation the final answer is (3pi/4)+1
 
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  • #5
Amaelle said:
thanks a lot but i think you did a miscalculation the final answer is (3pi/4)+1

Whoops... you're right! I thought you just wanted the area :doh:
 

1. What is a double integral?

A double integral is a mathematical concept used in calculus to find the area under a two-dimensional surface. It involves integrating a function over a region in a two-dimensional space.

2. What are some common problems with double integrals?

Some common problems with double integrals include choosing the correct limits of integration, dealing with improper integrals, and determining the correct order of integration.

3. How do you solve a double integral?

To solve a double integral, you first need to determine the limits of integration for both variables. Then, you can use the appropriate integration techniques, such as substitution or integration by parts, to evaluate the integral.

4. What are some real-world applications of double integrals?

Double integrals have many real-world applications, such as calculating the volume of a three-dimensional object, finding the center of mass of an object, and determining the average value of a function over a two-dimensional region.

5. How can I check if my double integral is correct?

You can check your double integral by using various methods, such as graphing the region of integration and comparing it to the original function, using symmetry properties, or evaluating the integral using different orders of integration.

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