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
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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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Likes etotheipi
  • #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 type of mathematical operation that is used to calculate the area or volume of a two-dimensional or three-dimensional region. It involves integrating a function over a specific region in the coordinate plane.

2. What is the purpose of a double integral?

The purpose of a double integral is to find the area under a curve or the volume of a solid in two or three dimensions. It is often used in physics, engineering, and other fields to solve problems involving surface area, mass, and other physical quantities.

3. How do you solve a problem with a double integral?

To solve a problem with a double integral, you first need to identify the region of integration and set up the integral using the appropriate limits of integration. Then, you need to evaluate the integral using techniques such as substitution, integration by parts, or using a table of integrals. Finally, you can interpret the result in the context of the problem.

4. What are the common challenges when working with double integrals?

One common challenge when working with double integrals is setting up the limits of integration correctly. This often requires a good understanding of the geometry of the region being integrated. Another challenge is evaluating the integral, which can be a complex and time-consuming process. Additionally, working with functions that have multiple variables can also be challenging.

5. How can double integrals be applied in real life?

Double integrals have many applications in real life, particularly in fields such as physics, engineering, and economics. For example, they can be used to calculate the volume of a 3D object, the surface area of a curved object, or the mass of a 2D or 3D system. They can also be used to find the average value of a function over a certain region, which is useful in statistics and economics.

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