Convert the integral into polar coordinates

[inner08]

Homework Statement

Convert the following integrals into polar coordinates and then calculate them.

a) int(0 , 2^(1/2)) int(y, [(4-y^2)^1/2]) xydxdy .

Homework Equations

x = rcostheta
y = rsintheta
r = (x^2 + y^2)^(1/2)

The Attempt at a Solution

Would it simply be:

int(0 , 2^(1/2)) int(rsintheta, [(4-(rsintheta)^2)^1/2]) r^2sinthetacostheta r dr dtheta

Is that all I have to do to convert it? I know how to integrate..I just need a bit of help with converting it into polar coordinates.

Thanks.

 

[dynamicsolo]

Homework Statement

Convert the following integrals into polar coordinates and then calculate them.

a) int(0 , 2^(1/2)) int(y, [(4-y^2)^1/2]) xydxdy .
Would it simply be:

int(0 , 2^(1/2)) int(rsintheta, [(4-(rsintheta)^2)^1/2]) r^2sinthetacostheta r dr dtheta

Is that all I have to do to convert it? I know how to integrate..I just need a bit of help with converting it into polar coordinates.

It looks to me like almost everything is converted. If I'm reading your statement correctly, your integration limits on y are 0 to sqrt(2). Won't that need to be converted also? (It might be a good idea to draw a picture of the integration region to see how to express it in polar coordinates. That may also make it clearer how to deal with the other set of integration limits.)

 

[HallsofIvy]

No, no, no! You have to change the limits of integration to polar coordinates as well. Draw a picture. x= y is a straight line. $x= (1-y^2)^{1/2}= \sqrt{1- y^2}$ is the right half of the unit circle $x^2+ y^2= 4$. You are integrating over the eighth of a circle between the x-axis and the line y= x. What are the upper and lower limits of r in that region? What are the upper and lower limits of $\theta$? You should find that the limits of integration are all constants!

The "No, no, no" was a response to inner08, not to dynamicsolo, who got in just before me.

 

[inner08]

Ok. I drew a graph. From looking at it, r would be from 0 to 1 and theta from 0 to pi/2?

By the way, shouldn't x=y and x=(4-y^2)^1/2 instead of x=(1-y^2)^1/2? If that is the case, r would be between 0 and 2 I think.

 

[D H]

How did you get r from 0 to 1 and theta from 0 to pi/2? Regarding the latter, pi/2 corresponds to an angle of 90⁰, or the y-axis. The line y=x represents a different angle.

Regarding the former, you have y alone going from 0 to $\surd2$. In other words, y just by itself goes outside the unit circle. BTW, the unit circle is $x^2+y^2equiv1$. You need to find some larger circle than the unit circle.

 

[HallsofIvy]

What is the radius of the circle? Yes, r starts at 0 but it has to go all the way out to the circle.

Now look at $\theta$. $\theta= 0$ to $\pi/2$ goes from the x-axis to the y-axis. Yes, you want to go up to the y-axis but you don't want to start at the x-axis, you want to start at the line y= x. What is $\theta$ for that?

 

[D H]

Yes, you want to go up to the y-axis but you don't want to start at the x-axis, you want to start at the line y= x.

You do not want to go the the y-axis. That corresponds to starting from x=0. The x integration starts at the line y=x and goes to the point on the circle that is at a height y above the x axis. Think of the x integration as following a line segment parallel to the x axis.

 

[inner08]

Ok, so theta is pi/4 so its from theta is from (0,pi/4). As for r, I'm still a bit confused. You said it has to go out of the circle so can I pick whatever I want like 3 or 4?..or how do I find it?

Thanks for being patient with me.

 

[D H]

Method 1: Sketch out the integration region on some graph paper. The circle's radius should be obvious.

Method 2: You already know its a circle, and you already know that it is the upper range if the x integration that hits the circle. You know the range of the y integration. What does this tell you about the circle's radius?

 

[HallsofIvy]

You do not want to go the the y-axis. That corresponds to starting from x=0. The x integration starts at the line y=x and goes to the point on the circle that is at a height y above the x axis. Think of the x integration as following a line segment parallel to the x axis.

I misread the problem! Didn't notice that it was integrating with respect to x first, then y. Guess I am too used to it being the other way around.

 

[HallsofIvy]

Ok, so theta is pi/4 so its from theta is from (0,pi/4). As for r, I'm still a bit confused. You said it has to go out of the circle so can I pick whatever I want like 3 or 4?..or how do I find it?

Thanks for being patient with me.

No, you cannot just "pick" whatever number you like!
The original problem said that the upper limit on the x-integral was x= [(4-y^2)^1/2]
Squaring both sides, $x^2= 4- y^2$ so $x^2+ y^2= 4$. That's the equation of a circle. What is the radius of that circle?