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Math Help - convert to polar and integrate question. (Hints appreciated)

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    Question 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?
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  2. #2
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    Quote Originally Posted by EuptothePiI1 View Post
    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 x=\sqrt{6} but also remember that x=r\cos(\theta) combining these two we get

    \displaystyle r\cos(\theta)=\sqrt{6} \iff r=\frac{\sqrt{6}}{\cos(\theta)}
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  3. #3
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    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?
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    Quote Originally Posted by EuptothePiI1 View Post
    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 \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
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