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Math Help - Finding the area of a circle using an integral

  1. #1
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    Finding the area of a circle using an integral

    I thought it would be fun to see if I could find the area of a circle with a radius of 5 by taking the integral of the positive form of a circle equation and multiplying it by two (since I believe the area of the bottom half is equal to the area of the upper half, or I should be able to take a definite integral from 0 to 5 and multiply it by 4). It turns out that integrating it was not as easy as I thought it would be.

    \int{\sqrt{25-x^2}}\text{ }dx

    I tried u-replacement, even though it looks like nothing will cancel.

    \int{\sqrt{u}}\text{ }dx

    u = 25-x^2

    \frac{du}{dx}=-2x

    du = -2x*dx

    dx = -\frac{1}{2x}*du

    \int{\sqrt{u}}*\frac{1}{2x}\text{ }du

    u = 25-x^2

    x = \sqrt{25-u}

    \int{\sqrt{u}}*\frac{1}{2\sqrt{25-u}}\text{ }du

    I don't think there's anything I can do with this.

    It looks like the equation fits one of the inverse trigonomic integrals, but it's not a fraction.

    Next, I try to integrate by parts. I can't really think of anything else.

    \int{\sqrt{25-x^2}}\text{ }dx = \int{\sqrt{(5+x)}*\sqrt{(5-x)}}

    By the way, if there's something I should have seen here that would indicate to me that integrating by parts here will not work, let me know.

    u = \sqrt{5-x}

    v = \sqrt{5+x}

    \frac{du}{dx} = \frac{1}{2}(5-x)^{-\frac{1}{2}}  * -1

    \int{\sqrt{5+x}} = \frac{2}{3}(5+x)^{\frac{3}{2}}

    \int{\sqrt{(5+x)} * \sqrt{(5-x)}} =\sqrt{5-x} * \frac{2}{3}(5+x)^{\frac{3}{2}} - \int{\frac{2}{3}(5+x)^{\frac{3}{2}} * \frac{1}{2}(5-x)^{-\frac{1}{2}}  * -1}

    I don't think it helped. I could maybe integrate by parts again, but it would probably not make it any easier to solve. What could I do to solve this?
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  2. #2
    Super Member Random Variable's Avatar
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    EDIT: let x = 5sin(u)
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  3. #3
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    You must do something called a trigonometric substitution.

    For functions with the form \sqrt {a^2-x^2}, use x=a\sin\theta

    Therefore for your integral you substitute x=5\sin\theta

    I'll do the problem for you.

    2 \int_{-5}^5 \sqrt {25-x^2}dx

    Let x=5\sin\theta

    dx=5\cos\theta d\theta

    When x=5, \theta = \frac {\pi}{2}

    When x=-5, \theta = -\frac {\pi}{2}

    2 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \sqrt {25-25\sin^2\theta} *5\cos\theta d\theta

    2 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \sqrt {25(1-\sin^2\theta)}*5\cos\theta d\theta

    2 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} 5\cos\theta *5\cos\theta d\theta

    2 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} 25\cos^2\theta d\theta

    50 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \frac {1+\cos 2\theta} {2} d\theta

    25 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} 1+\cos 2\theta d\theta

    25 \bigg [x+\frac {\sin 2\theta}{2} \bigg ]_{-\frac {\pi}{2}}^{\frac {\pi}{2}}

    25\pi
    Last edited by chengbin; June 23rd 2009 at 11:58 AM.
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by Phire View Post
    I thought it would be fun to see if I could find the area of a circle with a radius of 5 by taking the integral of the positive form of a circle equation and multiplying it by two (since I believe the area of the bottom half is equal to the area of the upper half, or I should be able to take a definite integral from 0 to 5 and multiply it by 4). It turns out that integrating it was not as easy as I thought it would be.

    \int{\sqrt{25-x^2}}\text{ }dx

    I tried u-replacement, even though it looks like nothing will cancel.

    \int{\sqrt{u}}\text{ }dx

    u = 25-x^2

    \frac{du}{dx}=-2x

    du = -2x*dx

    dx = -\frac{1}{2x}*du

    \int{\sqrt{u}}*\frac{1}{2x}\text{ }du

    u = 25-x^2

    x = \sqrt{25-u}

    \int{\sqrt{u}}*\frac{1}{2\sqrt{25-u}}\text{ }du

    I don't think there's anything I can do with this.

    It looks like the equation fits one of the inverse trigonomic integrals, but it's not a fraction.

    Next, I try to integrate by parts. I can't really think of anything else.

    \int{\sqrt{25-x^2}}\text{ }dx = \int{\sqrt{(5+x)}*\sqrt{(5-x)}}

    By the way, if there's something I should have seen here that would indicate to me that integrating by parts here will not work, let me know.

    u = \sqrt{5-x}

    v = \sqrt{5+x}

    \frac{du}{dx} = \frac{1}{2}(5-x)^{-\frac{1}{2}} * -1

    \int{\sqrt{5+x}} = \frac{2}{3}(5+x)^{\frac{3}{2}}

    \int{\sqrt{(5+x)} * \sqrt{(5-x)}} =\sqrt{5-x} * \frac{2}{3}(5+x)^{\frac{3}{2}} - \int{\frac{2}{3}(5+x)^{\frac{3}{2}} * \frac{1}{2}(5-x)^{-\frac{1}{2}} * -1}

    I don't think it helped. I could maybe integrate by parts again, but it would probably not make it any easier to solve. What could I do to solve this?
    HELLO :


    but area of circle is :

    conclusion :
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by chengbin View Post

    2 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \sqrt {25(1-\sin^2\theta)}*5\cos\theta d\theta

    2 \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} 5\cos\theta *5\cos\theta d\theta
    as a cheap observation, don't ever forget the abolute value.

    in that case it's |\cos\theta|=\cos\theta, but since it's -\frac\pi2\le\theta\le\frac\pi2, which means that cosine is positive there, we can safety remove the absolute value bars.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    as a cheap observation, don't ever forget the abolute value.

    in that case it's |\cos\theta|=\cos\theta, but since it's -\frac\pi2\le\theta\le\frac\pi2, which means that cosine is positive there, we can safety remove the absolute value bars.
    Yes, that's correct, but that's being really technical.

    Would you lose marks if you don't state that in a test?
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  7. #7
    Math Engineering Student
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    I wouldn't consider that technical, it's just being a bit cautious, 'cause suppose you're solving the problem and by not adding the absolute value bars you could get a result which is not correct.

    Besides, dunno if that mark would go, dunno how strict is the professor.
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