# convert to polar and integrate question. (Hints appreciated)

• Dec 7th 2010, 05:35 PM
EuptothePiI1
convert to polar and integrate question. (Hints appreciated)
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.

Original:

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?
• Dec 7th 2010, 05:49 PM
TheEmptySet
Quote:

Originally Posted by EuptothePiI1
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.

Original:

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?

The vertical line on the right has the equation $\displaystyle x=\sqrt{6}$ but also remember that $\displaystyle x=r\cos(\theta)$ combining these two we get

$\displaystyle \displaystyle r\cos(\theta)=\sqrt{6} \iff r=\frac{\sqrt{6}}{\cos(\theta)}$
• Dec 7th 2010, 06:16 PM
EuptothePiI1
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?
• Dec 7th 2010, 06:23 PM
TheEmptySet
Quote:

Originally Posted by EuptothePiI1
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?

$\displaystyle \displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\int_{0}^{\frac{\sqr t{6}}{\cos(\theta)}}rdrd\theta \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2}\left( \frac{\sqrt{6}}{\cos(\theta}\right)^2d\theta=3\int _{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^{2}(\theta)d\th eta=3\tan(\theta)\bigg|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=6$