How do I set boundaries in a multiple definite integral equation?

[zaw054]

I've been trying to solve these types of problems for a while now and still can't figure out how to set boundaries for each integral in a multiple integral equation. One of the questions is as follows:

Compute the integral
$$\iint_{\Omega} (-2x + 4y - 2) \ \text{d}\Omega$$on the region $$\Omega = \left\{ (x,y) \in R^2; 4 \le x^2 + y^2 \le 16 \text{ and } y \ge 0 \text{ and } y \le x \right\}$$
And somehow, the boundaries of each integral to solve this equation are as follows :
$$\int_{0}^{\frac{\pi}{4}} \int_{2}^{4} \left( -2(r \cdot \cos \alpha) + 4 (r \cdot \sin \alpha - 2)\right) \cdot r \ \text{d}r \ \text{d}\alpha$$
I get everything except the part on defining the regions of each integral. Any explanation would be really helpful for me. Thanks.

 

[mario99]

I've been trying to solve these types of problems for a while now and still can't figure out how to set boundaries for each integral in a multiple integral equation. One of the questions is as follows:

Compute the integral
$$\iint_{\Omega} (-2x + 4y - 2) \ \text{d}\Omega$$on the region $$\Omega = \left\{ (x,y) \in R^2; 4 \le x^2 + y^2 \le 16 \text{ and } y \ge 0 \text{ and } y \le x \right\}$$
And somehow, the boundaries of each integral to solve this equation are as follows :
$$\int_{0}^{\frac{\pi}{4}} \int_{2}^{4} \left( -2(r \cdot \cos \alpha) + 4 (r \cdot \sin \alpha - 2)\right) \cdot r \ \text{d}r \ \text{d}\alpha$$
I get everything except the part on defining the regions of each integral. Any explanation would be really helpful for me. Thanks.

Imagine that you have $4$ functions and you want to find the area between them.

Two circles.
$x^2 + y^2 = 4 = 2^2$
$x^2 + y^2 = 16 = 4^2$

And two lines
$y = 0$
$y = x$

Basically, you need to find the area between the two circles going from $y = 0$ to $y = x$.


You want to find the green area. Therefore, in polar coordinate, it is like going $45^{o}$ degrees counterclockwise and from radius $r = 2$ to radius $r = 4$.

 

[mario99]

The green area looks complicated in the $xy$-plane. Why not move it to the $r\theta$-plane?



Finding this area is very easy. This is why you are using the polar coordinate. It converts the complicated area in the $xy$-plane to a (simple) rectangle area in the $r\theta$-plane which can be found easily. Here, you are integrating only one region. Unlike the cartesian coordinate, you will have to divide the region into three areas, and find each one separately. It would be complicated, messy, and unnecessary to use it when we have a straightforward alternative.

 

[zaw054]

The green area looks complicated in the $xy$-plane. Why not move it to the $r\theta$-plane?

View attachment 37861

Finding this area is very easy. This is why you are using the polar coordinate. It converts the complicated area in the $xy$-plane to a (simple) rectangle area in the $r\theta$-plane which can be found easily. Here, you are integrating only one region. Unlike the cartesian coordinate, you will have to divide the region into three areas, and find each one separately. It would be complicated, messy, and unnecessary to use it when we have a straightforward alternative.

Thank you very much for taking the time to answe my question. This explanation was very helpful to me. I still need to work on these problems more to get a better understanding tho. Thanks again!

 

[mario99]

Thank you very much for taking the time to answe my question. This explanation was very helpful to me. I still need to work on these problems more to get a better understanding tho. Thanks again!

You are welcome.