I was able to solve the integral below however when I draw the region and attempt to switch to polar I can't figure out what to set the upper and lower limits of r at.
integrate 1 dydx where x = 0 to sqrt(6) and y = -x to x
The solution is 6.
However, I must convert to polar for this exercise.
After I drew the region. (A large triangle that can be viewed as two separate triangles. Both with legs of length sqrt(6) and hypontenuse of sqrt(12). These triangles would meet at the x-axis and be in quadrants I and IV.) I was able to determine that theta went from -pi/4 to pi/4. However I cannot figure out what to do for the limits of integration with r.
So the new integral needs to be r drdt (let t be theta) r = ? to ? and t = -pi/4 to pi/4. Any hints on where to go from here?
Howdy. I tried to implement that into my solution but I must have messed up in another section of my process. I attempted to solve the integral using dr dt where r is from 0 to sqrt(6)/cos(t) and t is from -pi/4 to pi/4. This leaves us with 2sec^3(t) from -pi/4 to pi/4 at the end which gives us zero. I feel as if I missed something of great importance. Can you see where I went wrong?