Double Integrals in Polar Coordinates

[Daniel_Feldman]

I need to evaluate the double integral of (1/2)sqrt(2x^2+2y^2)dA for the region R, where R is the region given by:

with a=5 and b=8. I am to assume that the curved boundary of the figure is circular.


I believe this should be set up in polar coordinates. So I think that my inner limits would be 5 and 8, and then my outer limits would be tan^-1(5/8) and pi-tan^-1(5/8). However, evaluating this gets me the wrong answer. Can someone tell me where I'm going wrong?

 

[Count Iblis]

Did you use the correct integration measure:

dA = r dr dtheta ?

 

[Count Iblis]

Also, from the figure the integration limits for the angle (taken relative to the negative x-axis, clockwise is positive) should be from arctan(5/8) to
pi/2 - arctan(5/8). The radius is from zero to sqrt[5^2 + 8^2].

 

[soroban]

Hello, Daniel!

I need to evaluate: \(\displaystyle \L\int\int\)\(\displaystyle \frac{1}{2}\sqrt{2x^2\,+\,2y^2}\,dA\)
for the region \(\displaystyle R\), where \(\displaystyle R\) is the region given by:

with \(\displaystyle a\,=\,5\) and \(\displaystyle b\,=\,8.\)
I am to assume that the curved boundary of the figure is circular.

I believe this should be set up in polar coordinates.
So I think that my inner limits would be 5 and 8, . . . . no
and then my outer limits would be \(\displaystyle \tan^{-1}\left(\frac{5}{8}\right)\) and \(\displaystyle \pi\,-\,\tan^{-1}\left(\frac{5}{8}\right).\) . . . . no

Think . . . \(\displaystyle \tan^{-1}\left(\frac{5}{8}\right)\) is in Quadrant I.

First, convert the function to polar coordinates . . .
. . \(\displaystyle \frac{1}{2}\sqrt{2x^2\,+\,2y^2} \:=\:\frac{1}{2}\sqrt{2(x^2\,+\,y^2)}\;=\;\frac{1}{2}\sqrt{2r^2}\;=\;\frac{\sqrt{2}}{2}r\)

The radius of the circle is: \(\displaystyle \,r\:=\:\sqrt{8^2\,+\,5^2}\:=\:\sqrt{89}\)
. . The limits are: \(\displaystyle \:r\,=\,0\) to \(\displaystyle r\,=\,\sqrt{89}\)

The two angles are: .\(\displaystyle \begin{array}{cc}\theta_1\:=\:\tan^{-1}\left(-\frac{8}{5}\right) \\ \theta_2\:=\:\tan^{-1}\left(-\frac{5}{8}\right) \end{array}\)


The integral is: \(\displaystyle \L\int^{\;\;\;\;\theta_2}_{\theta_1} \int^{\;\;\;\;\sqrt{89}}_0 \frac{\sqrt{2}}{2}r\,dr\,d\theta\)

 

[galactus]

Don't forget that extra r in polar coordinates. Wouldn't it be:

\(\displaystyle \L\\\int\int\frac{\sqrt{2}}{2}r^{2}drd{\theta}\)?.

Just checkin'.