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Math Help - i have no idea where to even begin with this crazy integral question about unit circl

  1. #1
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    i have no idea where to even begin with this crazy integral question about unit circl

    we have already prooved that pi is the area of the unit disk and pi=2(integral from -1 to 1) square root(1-x^2) dx...use properties of the integral to compute the following in terms of pi
    a. (integral from -3 to 3) square root (9-x^2) dx
    b. (integral from 2 to 0) square root (1- 1/4 x^2) dx
    c. (integral from -2 to 2) (x-3) square root (4-x^2) dx

    im sorry about the poor notation, but i have no other idea how to write it.....i appreciate any help whatsoeever and thanks in advance.
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  2. #2
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    Quote Originally Posted by twostep08 View Post
    we have already prooved that pi is the area of the unit disk and pi=2(integral from -1 to 1) square root(1-x^2) dx...use properties of the integral to compute the following in terms of pi
    a. (integral from -3 to 3) square root (9-x^2) dx
    b. (integral from 2 to 0) square root (1- 1/4 x^2) dx
    c. (integral from -2 to 2) (x-3) square root (4-x^2) dx

    im sorry about the poor notation, but i have no other idea how to write it.....i appreciate any help whatsoeever and thanks in advance.
    You have \pi = 2 \int_{-1}^{1} \sqrt{1 - x^2} \, dx = 4 \int_{0}^{1} \sqrt{1 - x^2} \, dx.

    a. Substitute u = \frac{x}{3}.


    b. \int_{2}^{0} \sqrt{1 -\frac{x^2}{4}} \, dx = - \int_{0}^{2} \sqrt{1 -\frac{x^2}{4}} \, dx. Substitute u = \frac{x}{2}.


    c. \int_{-2}^{2} (x - 3) \sqrt{4 - x^2} \, dx = \int_{-2}^{2} x \sqrt{4 - x^2} \, dx - 3 \int_{-2}^{2}\sqrt{4 - x^2} \, dx = I_1 - 3 I_2.

    To calculate I_1 substitute u = 4 - x^2. To calculate I_2, think about b.
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  3. #3
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    oh sorry i didnt mention before, i cant use u substitution...
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