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Thread: finding the area using double integrals

  1. #1
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    finding the area using double integrals

    Given $\displaystyle r=\cos(\theta)$ and $\displaystyle r=\sin(\theta)$ 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

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

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

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

    but this gives me $\displaystyle \frac{1}{8}$, whereas the solution in the back of the book tells me it's supposed to be $\displaystyle \frac{\pi-2}{8}$
    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:

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

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

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


    Divide through by $\displaystyle \cos{(\theta)} $


    $\displaystyle 1 = \tan{(\theta)} $

    (Answer should be $\displaystyle 0 \leq r \leq \frac{\sqrt{2}}{2} $)
    Last edited by Mush; Jan 16th 2009 at 09:58 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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    Quote Originally Posted by lllll View Post
    Given $\displaystyle r=\cos(\theta)$ and $\displaystyle r=\sin(\theta)$ 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

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

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

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

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

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

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

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

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

    $\displaystyle \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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    I must have made a mistake in my calculation, but I'm still no were near my final answer.

    I've tried :

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

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

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

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

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

    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
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    Quote Originally Posted by lllll View Post
    Given $\displaystyle r=\cos(\theta)$ and $\displaystyle r=\sin(\theta)$ 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

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

    but this gives me $\displaystyle \frac{1}{8}$, whereas the solution in the back of the book tells me it's supposed to be $\displaystyle \frac{\pi-2}{8}$
    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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