cartesian integral to polar integral

[mindy88]

The question is change the cartesian integral into an equivalent polar integral. Then evaluate the polar integral

it's the integral of 0 to 6, the integral of o to y of x dx dy

the answer i got it
the integral from 0 to pi/4, the integral of 0 to blank of r cos theta r dr d(theta)

i don't know what the blank is
i think i set x=y and somehow find the value of r from that, to find out what the blank is, but i don't know how.

 

[galactus]

Try:

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{4}}\int_{0}^{12sin{\theta}}[r^{2}cos{\theta}]drd{\theta}\)

Since \(\displaystyle \L\\r=\sqrt{6^{2}+6^{2}}=6\sqrt{2}\)

We have \(\displaystyle \L\\a\cdot{sin(\frac{\pi}{4})}=6\sqrt{2}\)

\(\displaystyle \L\\a=12\)


EDIT: I made a mistake before. You are correct about the Pi/4. \(\displaystyle y=rsin{\theta}, \;\ x=rcos{\theta}\)

 

[mindy88]

i think i see why it's from 0 to 6 but why is it pi/6?