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Math Help - Find the area of the region bounded

  1. #1
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    Find the area of the region bounded

    Question : Find the area bounded by \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 and above the x-axis
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question : Find the area bounded by \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 and above the x-axis

    By symmetry wrt the x-axis, this is just half the area of the canonical ellipse with semi-axis a,b , i.e. \frac{\pi ab}{2}, but of course you need to prove this with integral calculus, so:

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Longrightarrow a^2y^2=b^2(a^2-x^2)\Longrightarrow y=b\sqrt{1-\left(\frac{x}{a}\right)^2} , so you need the integral \int_{-a}^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx=2b\int_0^a\sqrt{1-\left(\frac{x}{a}\right)^2}dx , again by symmetry, this time wrt the y-axis.

    You may want to try the trigonometric substitution x=a\sin t\,,\,0\le t\le \frac{\pi}{2} , to make things easier.

    Tonio
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    Is this correct?

    Quote Originally Posted by tonio View Post
    By symmetry wrt the x-axis, this is just half the area of the canonical ellipse with semi-axis a,b , i.e. \frac{\pi ab}{2}, but of course you need to prove this with integral calculus, so:

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Longrightarrow a^2y^2=b^2(a^2-x^2)\Longrightarrow y=b\sqrt{1-\left(\frac{x}{a}\right)^2} , so you need the integral \int_{-a}^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx=2b\int_0^a\sqrt{1-\left(\frac{x}{a}\right)^2}dx , again by symmetry, this time wrt the y-axis.

    You may want to try the trigonometric substitution x=a\sin t\,,\,0\le t\le \frac{\pi}{2} , to make things easier.

    Tonio

    Taking x \ = \ a sint as 0 \le t < \frac{ \pi}{2}

    = 2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \left( \frac{a \ sint}{a} \right)^2}

    = 2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sint )^2}

    = 2b \int_{0}^{\frac{\pi}{2}} 1 - ( sint )^2

    = 2b \int_{0}^{\frac{\pi}{2}} 1 - sint

    = b \pi...............Is this corect???
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  4. #4
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    your set up should look like: 2 integral (from 0 to a) of sqrt(b^2 - (b^2)(x^2)/a^2) dx.

    you do not need a trig substitution, simply factor out a b^2/a^2 term inside the square root and you will get sqrt[(b^2/a^2) (a^2 - x^2)]. sqrt(b^2/a^2) becomes b/a so you can factor that constant out of the integral. so now the integral should look like 2b/a integral (from 0 to a) of sqrt(a^2 - x^2) dx. now look at the integral, and geometrically it just means the area of a circle of radius "a" and since the limit is from 0 to a, the integral only wants 1/4 of the whole circle. so
    integral (from 0 to a) of sqrt(a^2 - x^2) dx is just (pi)(a^2)/4 since it's 1/4 of the area of the whole circle. then with the constant outside of the integral earlier, you get (2b/a)((pi)(a^2)/4)) = (pi)(a)(b)/2 and since the area of an ellipse is pi(a)(b), your result makes sense because you only wanted the top half of the ellipse.
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  5. #5
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    please look at tonio post

    Quote Originally Posted by oblixps View Post
    your set up should look like: 2 integral (from 0 to a) of sqrt(b^2 - (b^2)(x^2)/a^2) dx.

    you do not need a trig substitution, simply factor out a b^2/a^2 term inside the square root and you will get sqrt[(b^2/a^2) (a^2 - x^2)]. sqrt(b^2/a^2) becomes b/a so you can factor that constant out of the integral. so now the integral should look like 2b/a integral (from 0 to a) of sqrt(a^2 - x^2) dx. now look at the integral, and geometrically it just means the area of a circle of radius "a" and since the limit is from 0 to a, the integral only wants 1/4 of the whole circle. so
    integral (from 0 to a) of sqrt(a^2 - x^2) dx is just (pi)(a^2)/4 since it's 1/4 of the area of the whole circle. then with the constant outside of the integral earlier, you get (2b/a)((pi)(a^2)/4)) = (pi)(a)(b)/2 and since the area of an ellipse is pi(a)(b), your result makes sense because you only wanted the top half of the ellipse.

    the reason i did the trig substitution is because the post above directed me to do it
    please look at tonio's post above and let me know whether to do it or no???
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  6. #6
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    I wasn't saying that you shouldn't do the trig substitution. I'm just offering an alternative way of doing the problem which may be easier. you can do the problem either way. all that matters is that you get the correct answer.
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  7. #7
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    Ok i interpreted it wrongly

    Quote Originally Posted by oblixps View Post
    I wasn't saying that you shouldn't do the trig substitution. I'm just offering an alternative way of doing the problem which may be easier. you can do the problem either way. all that matters is that you get the correct answer.

    Oh !!! Ok
    thanks mite
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  8. #8
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    Quote Originally Posted by zorro View Post
    Taking x \ = \ a sint as 0 \le t < \frac{ \pi}{2}

    = 2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \left( \frac{a \ sint}{a} \right)^2}

    = 2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sint )^2}

    = 2b \int_{0}^{\frac{\pi}{2}} 1 - ( sint )^2

    = 2b \int_{0}^{\frac{\pi}{2}} 1 - sint

    = b \pi...............Is this corect???
    you forgot a few things
    x=asinz
    dx=acosz dz
    2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sinz )^2}acoszdz
    2b \int_{0}^{\frac{\pi}{2}} \sqrt{(cosz)^2}acoszdz
    2b \int_{0}^{\frac{\pi}{2}} cosz a coszdz
    2b \int_{0}^{\frac{\pi}{2}} a(cosz)^2
    2ab \int_{0}^{\frac{\pi}{2}} (\frac{1+cos2z}{2}
    ab [\int_{0}^{\frac{\pi}{2}} (1)+\int_{0}^{\frac{\pi}{2}}cos2z]
    ab [\frac{\pi}{2}+\frac{1}{2}(sin(\frac{2\pi}{2})-0)]
    ab [\frac{\pi}{2}]=\frac{ab\pi}{2}
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