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Thread: [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.
    $\displaystyle A=\pi\int_a^{b}A(x)dx$

    where $\displaystyle A(x)$ is to be determined by finding in this case what the radius would be in $\displaystyle A=\frac{1}{2}\pi{r^2}$
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    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 $\displaystyle \int_a^b A(x)\,dx$, where $\displaystyle A(x)$ is the area of the cross section at $\displaystyle x$. Since the cross section is a quarter circle, its area is $\displaystyle \frac14\pi r^2$. We can find $\displaystyle r$ to be the length of the chord in the base perpendicular to the $\displaystyle x$-axis at any given value of $\displaystyle x$, so $\displaystyle r = 2\sqrt{4-x^2}$. Since the base is a circle of radius 2 centered at the origin, the cross sections are taken for $\displaystyle x\in[-2,2]$. The total area is then

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

    $\displaystyle =\pi\int_{-2}^2\left(4 - x^2\right)\,dx$
    Last edited by Reckoner; Jun 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
    $\displaystyle A=\pi\int_a^{b}A(x)dx$

    where $\displaystyle A(x)$ is to be determined by finding in this case what the radius would be in $\displaystyle A=\frac{1}{2}\pi{r^2}$
    Almost. We are dealing with cross sections, not solids of revolution, so you do not need the $\displaystyle \pi$ in front of the integral. Also, the cross sections are quarter circles, so the formula we want is $\displaystyle A = \frac14\pi r^2$. 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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