# Polar Coordinates problem

Hi all,
I have a question on a homework assignment that says I have to evaluate a given integral by converting to polar coordinates. It gives us a sum of two integrals that we can combine into 1 integral by converting to polar coordinates. Im not having trouble getting down to one integral, but my issue is when i actually try to integrate. I know x = rcos(theta) and y = rsin(theta). The function was f(x,y) = xy. So then when I convert to polar coordinates, it becomes the double integral of (rsin(theta)) * (rcos(theta) r dr d theta right? I don't need help with the limits of integration, thats why I am not including them. So I integrate with respect to r first, and get (r^4/4)*(sin(theta) * cos(theta)), for some interval values, dtheta. Evaluate that integral with the appropriate interval values, then I try to integrate with respect to theta now, and I end up with the integral of sin(theta) * cos(theta) dtheta, and I don't know how to integrate this :( I think I set it up right, but I'm confused to how to integrate sin(theta)*cos(theta).

I hope I explained this clearly enough....

Lisa
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Commented:
Since I do not have the limits of integration I will assume your integration to date is correct (although it seems a a bit odd), to do the final bit

integral ( sin(theta) cos(theta) dtheta )

use the identity sin(2theta) = 2sin(theta) cos(theta)  so that

sin(theta) cos(theta) =  sin(2theta)/2

which I am sure you can integrate (if not let me know)

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What concerns me about the initial bit you have shown that if f(xy)=xy then f(r,theta) is in fact f(r,theta) = r² sin(theta)cos(theta) with the original integral being the integral of over the same region, so that the limits change with one set being fuctions of the other variable.
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Author Commented:
GwynforWeb,
Thanks for your response. Well I could include the limits of integration if you like.

Here is the whole problem:

A. Sketch and shade the region of integration for the sum of the two integrals:

(Integral from 0 to (5sqrt(2))/2 integral from 0 to x of xy dydx) + (Integral from (5sqrt(2))/2 to 5 integral from 0 to sqrt(25-x²) of xy dydx)
Obviously I sketched it, by knowing:
0 <= y <= x
0 <= x < (5sqrt(2))/2
0 <= y <= sqrt(25-x²)
(5sqrt(2))/2 <= x <= 5

B. Rewrite the sum of the two integrals in part A as a single iterated integral, by converting to polar coordinates.

I get: Integral from 0 to pi/4 integral from 0 to 5 of (rsin(theta)*rcos(theta) r drdtheta), which simplifies to:
Integral from 0 to pi/4 integral from 0 to 5 of (r^3 (sin(theta)*cos(theta)drdtheta)

C. Evaluate this integral.
Which goes to the same question as I had before.

I hope this can help clarify things. I'm not to sure if I understand where you are getting f(r,theta) = r² sin(theta)cos(theta) as you said before?

Thanks!
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Commented:
Hmm, there is some thing not right here, bit rushed for time here so will look properly if I have time.

But the relationship between polar and cartesian coord is

x=rcos(theta),  y=rsin(theta)

so for a function f(x,y) = xy then at some point in the plane  (r, theta) you merely substitue for x and y into f(x,y) = xy giving at (r, theta) that

f(r,theta) =rcos(theta)rsin(theta)

=r² sin(theta)cos(theta)

= r²( sin(2theta))/2

(ie the product of the x and y values corressponding to the point (r,theta) in the plane, do not confuse this process with the transforms you make for 1D-integrals)

Any way
integral ( sin(theta) cos(theta) dtheta )=integral ( sin(2theta)/2 ) dtheta
=-cos(2theta)/4

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Author Commented:
Hi GwynforWeb,
I think you are missing the extra factor of r that multiplies drdtheta.

Brief explanation if you are interested...
Points are described as (r, theta) rather than (x,y), and the area of a tiny region is not dxdy but rather rdrdtheta. This is because the radian measure of angle has the property that the length of a piece of a circle radius r sweeping out an angle of change in theta is r times the change in theta. A small wedge in which the radius changes from r to r + change in r and the angle changes from theta to change in theta has area that is approximately equal to change in r times r times change in theta. The extra r-factor comes about because one side of the wedge has length of about r times the change in theta rather than just change in theta. And computing double integrals in polar coordinates is done by iterating two single integrals.

Hope that helped explained where I'm getting that extra "r" from, but I guess regardless of that issue, I will still need to use the double angle formulas to integrate.

Thanks,
Aaliyah
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Commented:
Aaliyah9152,
I know what you are saying, but I assure you  r² sin(theta)cos(theta) in polar coords is the same surface as xy in cartesian coords. The integral is the volume under the surface for the region defined by the limits. There are different ways of doing this and I see what you are doing, it is fine.

GfW

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Commented:
hey i think i know it
what u did is right just put

sin (theta) * cos (theta) = (0.5)* sin (2* theta)

therefore you have integral of 0.5* sin (2* theta) d theta

now note : integral sin(theta) = -cos (theta)
hence integral sin (k*theta) = - (cos(theta))/k
(you may diffrenciate the LHS and verify for yourself)

therefore experssion (1) evaluates to -0.25 * cos (2*theta) with limits ranging from 0 to pi/4
which is -0.25*( cos (pi/2) - cos (0))
=>        -0.25*( 0 - (1))
=>        -0.25 * -1
= 0.25

i hope this is what you were looking for 'cos i am in great need of points so kindly award them soon

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Author Commented:
Hi again,

darpangoel - thanks for response as well. Just one question, when you evaluated the integral from 0 to pi/4, you placed pi/2 into the equation. Was that a mistake, or where did you pull that pi/2 from?

I will up the points and split them between the two of you.

Thanks!
Aaliyah
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Commented:
EXCUSE ME!!  but I gave that integral already,  ie i wrote

" Any way
integral ( sin(theta) cos(theta) dtheta )=integral ( sin(2theta)/2 ) dtheta
=-cos(2theta)/4 "

which when evaluated between 0 and pi/4 is

0.25cos(2*pi/4) - 0.25cos(2*0) = -0.25

GfW  :-|
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Commented:
look at my first response as well.
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Author Commented:
GwynforWeb,
My apologies! I have overlooked that. I am very sorry. I will post another question so I can award you more points.

My apologies again. =(

Aaliyah
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