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Math Help - [SOLVED] Volume

  1. #1
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    [SOLVED] Volume

    A solid has, as its base, the circular region in the xy-plane bounded by the graph of x^2+y^2=4. FInd the volume of the solid if every cross section by a plane perpendicular to the x-axis is a quarter circle with one of its radii in the base.

    I really can't picture this. I tried to draw it out, but I'm unable to get the right answer.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chukie View Post
    A solid has, as its base, the circular region in the xy-plane bounded by the graph of x^2+y^2=4. FInd the volume of the solid if every cross section by a plane perpendicular to the x-axis is a quarter circle with one of its radii in the base.

    I really can't picture this. I tried to draw it out, but I'm unable to get the right answer.
    A=\pi\int_a^{b}A(x)dx

    where A(x) is to be determined by finding in this case what the radius would be in A=\frac{1}{2}\pi{r^2}
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  3. #3
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by chukie View Post
    A solid has, as its base, the circular region in the xy-plane bounded by the graph of x^2+y^2=4. FInd the volume of the solid if every cross section by a plane perpendicular to the x-axis is a quarter circle with one of its radii in the base.

    I really can't picture this. I tried to draw it out, but I'm unable to get the right answer.
    The area is just \int_a^b A(x)\,dx, where A(x) is the area of the cross section at x. Since the cross section is a quarter circle, its area is \frac14\pi r^2. We can find r to be the length of the chord in the base perpendicular to the x-axis at any given value of x, so r = 2\sqrt{4-x^2}. Since the base is a circle of radius 2 centered at the origin, the cross sections are taken for x\in[-2,2]. The total area is then

    \int_{-2}^2\frac14\pi\left(2\sqrt{4 - x^2}\right)^2\,dx

    =\pi\int_{-2}^2\left(4 - x^2\right)\,dx
    Last edited by Reckoner; June 29th 2008 at 05:48 PM. Reason: Minor correction
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  4. #4
    MHF Contributor Reckoner's Avatar
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    chukie, I made up a quick graph of this since you said that you had difficulty visualizing it.



    Note that we have a circular base, and along the left half of it lies the "center" of our quarter-circles (what I mean to say is, the center of the circles that the quarter-circles belong to), and the radius of these circles extend from one side of the base to the other.

    Quote Originally Posted by Mathstud28 View Post
    A=\pi\int_a^{b}A(x)dx

    where A(x) is to be determined by finding in this case what the radius would be in A=\frac{1}{2}\pi{r^2}
    Almost. We are dealing with cross sections, not solids of revolution, so you do not need the \pi in front of the integral. Also, the cross sections are quarter circles, so the formula we want is A = \frac14\pi r^2. Not sure how I missed this, since it says you posted before me.
    Attached Thumbnails Attached Thumbnails [SOLVED] Volume-graph_20080629.png   [SOLVED] Volume-spacer.png  
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