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Math Help - finding the area using double integrals

  1. #1
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    finding the area using double integrals

    Given r=\cos(\theta) and r=\sin(\theta) find the area inside both circle.

    Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

    \int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta

    but this gives me \frac{1}{8}, whereas the solution in the back of the book tells me it's supposed to be \frac{\pi-2}{8}
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  2. #2
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    Quote Originally Posted by lllll View Post
    Given r=\cos(\theta) and r=\sin(\theta) find the area inside both circle.

    Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

    \int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta

    but this gives me \frac{1}{8}, whereas the solution in the back of the book tells me it's supposed to be \frac{\pi-2}{8}
    There are two points of intersection in that quadrant. One of them is clearly 0, and the other angle at which they intersect is a solution to the equation:

     \cos{(\theta)} = \sin{(\theta)} .

    Solve that, then plug your solution into the original equations  r = \cos{(\theta)} or  r = \sin{(\theta)} . This will give you the REAL r limits.
    Hint...

     \cos{(\theta)} = \sin{(\theta)}


    Divide through by  \cos{(\theta)}


    1 = \tan{(\theta)}

    (Answer should be  0 \leq r \leq \frac{\sqrt{2}}{2} )
    Last edited by Mush; January 16th 2009 at 10:58 AM.
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  3. #3
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    I see that the they interest at those given points, but shouldn't my area be bounded by the given equations?
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  4. #4
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    Quote Originally Posted by lllll View Post
    Given r=\cos(\theta) and r=\sin(\theta) find the area inside both circle.

    Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

    \int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta

    but this gives me \frac{1}{8}, whereas the solution in the back of the book tells me it's supposed to be \frac{\pi-2}{8}
    BTW, how did you manage to get  \frac{1}{8} anyway? By my reckoning:

    \displaystyle \int_0^{\frac{\pi}{2}} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta

    \displaystyle = \int_0^{\frac{\pi}{2}} [\frac{r^2}{2}]_{\sin(\theta)}^{\cos(\theta)} \ d\theta

    \displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{2} -\frac{\sin^2{\theta}}{2} d\theta

    \displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2{\theta} - \sin^2{\theta}}{2}  d\theta

    \displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos{2\theta}}{2}  d\theta

    \displaystyle = \frac{1}{4} [\sin{2\theta}]_0^{\frac{\pi}{2}}

    \displaystyle = \frac{1}{4} (\sin{\pi} - \sin{0}) = 0

    Clearly your integral was not correct to begin with, but neither was your integration process!
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  5. #5
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    I must have made a mistake in my calculation, but I'm still no were near my final answer.

    I've tried :

    \int_0^{\pi/2} \int_0^{\sqrt{2}/2} tan(\theta) \cdot r \ dr \ d\theta  = undifined

    \int_0^{\pi/2} \int_0^{\sqrt{2}/2} r \ dr \ d\theta  =frac{1}{8} \pi

    \int_0^{\pi/2} \int_{\cos(\theta)}^{\sin(\theta)} tan(\theta) \cdot r \ dr \ d\theta  = undifended

    even changing \frac{\pi}{2} \rightarrow \pi  \ \mbox{or} \ 2\pi got me nowhere
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  6. #6
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    Try this:

    2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{cos  ({\theta})}rdrd{\theta}

    See why?. The region shaded in the graph is half the region needed, so we multiply by 2 since there is symmetry.
    Attached Thumbnails Attached Thumbnails finding the area using double integrals-cos-t-.jpg  
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  7. #7
    Senior Member DeMath's Avatar
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    Quote Originally Posted by lllll View Post
    Given r=\cos(\theta) and r=\sin(\theta) find the area inside both circle.

    Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

    \int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta

    but this gives me \frac{1}{8}, whereas the solution in the back of the book tells me it's supposed to be \frac{\pi-2}{8}
    This is a graphical representation of your problem in polar coordinates.



    I hope you will have better understanding of your problem.
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