# Math Help - Double Integration - polar co-ordinates

1. ## Double Integration - polar co-ordinates

I have to evaluate the integral between 0 and 2a of the integral between 0 and sqrt(2ax-x^2) of x^2 dxdy

I am supposed to draw a diagram and draw a radial slice and I have to convert to polar co-ordinates.

The only thing I know how to do is x becomes rcos(theta) and dxdy becomes r dr dtheta

Any help would be great thanks.

2. Originally Posted by jpquinn91
I have to evaluate the integral between 0 and 2a of the integral between 0 and sqrt(2ax-x^2) of x^2 dxdy

I am supposed to draw a diagram and draw a radial slice and I have to convert to polar co-ordinates.

The only thing I know how to do is x becomes rcos(theta) and dxdy becomes r dr dtheta

Any help would be great thanks.
The region of integration is the top half a circle of radius a and center (a,0). In terms of polar coords

$
\int_0^{\frac{\pi}{2}} \int_0^{2a \cos \theta} r^2 \cos^2 \theta r dr d \theta
$

3. Thanks for the help, could you please tell me how you know what to change the limits to and what the equation/centre of the circle is?

4. Originally Posted by jpquinn91
Thanks for the help, could you please tell me how you know what to change the limits to and what the equation/centre of the circle is?
You are going from a botton curve $y = 0$ to a top curve $y = \sqrt{2ax-x^2}$. For the latter, squaring both sides gives

$y^2 = 2ax-x^2 \;\text{or}\; x^2 -2ax + a^2 + y^2 = a^2 \; \text{or}\; (x-a)^2 + y^2 = a^2.$ (Just the top half).

In polar coords this becomes

$r^2 = 2ar \cos \theta$ or just $r = 2a \cos \theta$.

So we have a 1/2 circle centered at $(a,0)$ with radius $r = a$. In order to find limits we ask, "for what values of $\theta$ and $r$ will sweep out this half circle. Here $\theta$ will go from 0 to $\frac{\pi}{2}$and r from $r = 0$ to the outside curve $r = 2a \cos \theta$. That's how I got the limits.

5. OK, i think i've got the answer. Is 4a^4 right? I just don't understand where r^2 = 2arcos(theta) comes from.

6. Originally Posted by jpquinn91
OK, i think i've got the answer. Is 4a^4 right? I just don't understand where r^2 = 2arcos(theta) comes from.
Not sure how you got your answer. You might want to provide some more details so we can check. As for the the $r^2 = 2a r \cos \theta$ we have $
x^2 +y^2 = 2 a x
$
. Now $x^2+y^2 = r^2$ and $x = r \cos \theta$. Just substitute.

7. I evaluated the integral that you wrote down after changing to polar co-ords and got:

Int between Pi/2 & 0 of (cos^2(theta)*(2acos(4theta)))^4 /4 d theata

which is: Int between Pi/2 & 0 of (16a^4(cos^6(theta))) /4 d theta

= 4a^4(sin^6(theta)) , subs in values of theat

which is: 4a^4 - 0 = 4a^4 i think ?????

8. Originally Posted by jpquinn91
I evaluated the integral that you wrote down after changing to polar co-ords and got:

Int between Pi/2 & 0 of (cos^2(theta)*(2acos(4theta)))^4 /4 d theata

which is: Int between Pi/2 & 0 of (16a^4(cos^6(theta))) /4 d theta

= 4a^4(sin^6(theta)) , subs in values of theat

which is: 4a^4 - 0 = 4a^4 i think ?????
Problem $\int \cos^6 \theta d \theta \ne \sin^6 \theta + c$

9. is this right? Int cos^6(theta) = (9*sin(4*x)-4*sin(2*x)^3+48*sin(2*x)+60*x)/192

10. Originally Posted by jpquinn91
is this right? Int cos^6(theta) = (9*sin(4*x)-4*sin(2*x)^3+48*sin(2*x)+60*x)/192
Yep - Maple agrees.

11. Does evaluate mean find the area or just write down the integral seeing as I don't think I would be expected to use maple?

12. Originally Posted by jpquinn91
Does evaluate mean find the area or just write down the integral seeing as I don't think I would be expected to use maple?
I would say evaluate the integral.

13. So is 5/8 *Pi*a^4 correct? Can you do these double integrals on maple?

14. Originally Posted by jpquinn91
So is 5/8 *Pi*a^4 correct? Can you do these double integrals on maple?