can someone walk me through this to me please? polar coordinates confuse me. i know how to do double integrals, but not how to determine bounds within a polar and what to sub where.

Using polar coordinates, evaluate the integral ∫∫sin(x^2+y^2)dA where R is the region 16≤x^2+y^2≤81.

Question

can someone walk me through this to me please?  polar coordinates confuse me.  i know how to do double integrals, but not how to determine bounds within a polar and what to sub where.  

Using polar coordinates, evaluate the integral ∫∫sin(x^2+y^2)dA where R is the region 16≤x^2+y^2≤81.

zepdrix · Answer

Hey there :)
So recall that x^2+y^2 represents a circle, yah?

zepdrix · Answer

So the region that we're integrating over is the space between two circles.

zepdrix · Answer

|dw:1414227592454:dw|We want to find the area between these two circles.
Understand how I was able to come up with those radius values?

zepdrix · Answer

To convert to polar we use these:$$\Large\rm x=r \cos \theta$$$$\Large\rm y=r \sin \theta$$Recall that when you change from Cartesian to Polar, your differentials will pick up a factor of r,$$\Large\rm dA=\quad dy~dx=\quad r~dr~d\theta$$

anonymous · Answer

$$\int\limits_{0}^{2\pi}\int\limits_{6}^{81}\sin(r^2)drd$$??

anonymous · Answer

@zepdrix

zepdrix · Answer

One of our boundaries is this:$$\Large\rm 16\le x^2+y^2$$$$\Large\rm 4^2\le x^2+y^2$$Recall your circle equation:$$\Large\rm x^2+y^2=r^2$$We need to see that value squared so we can get a proper boundary for $\Large\rm r$.

zepdrix · Answer

So we want our radius to extend from r=4 out to r=9.

zepdrix · Answer

And, uh oh... you forgot your factor of r when you converted your differentials!!! :O

anonymous · Answer

ok, so it's $$\int\limits_{4}^{9}$$, not 16, 81