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Math Help - Volume of a revolving solid...

  1. #1
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    Volume of a revolving solid...

    I don't feel very confident with this concept.

    Suppose R is the region bounded by y = x2, x = 2, and y = 0

    1) Find the volume of a solid that is generated by revolving R about the x-axis. - I believe I did this one correctly, but I want to be sure. I got 32pi/5.
    2) (same) ... revolving R about the y-axis.
    - I feel the same about this one. I got 8pi.
    3) ... revolving R about y = -3.
    4) ... revolving R about x = 3.

    For 3 and 4, I've tried setting up the integrals a few different ways, but I don't think any of them are right.

    If someone could help me set these up or give me some tips & verify that the first two are correct, I'd be eternally grateful.
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  2. #2
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    Re: Volume of a revolving solid...

    1) $\displaystyle \begin{align*} V = \pi \int_0^2{ \left( x^2 \right) ^2 \, dx } \end{align*}$

    2) $\displaystyle \begin{align*} V = \pi \int_0^4{ 2^2 - \left( \sqrt{y} \right) ^2 \, dy } \end{align*}$

    3) $\displaystyle \begin{align*} V = \pi \int_0^2{ \left( x^2 + 3 \right) ^2 \, dx } \end{align*}$

    4) $\displaystyle \begin{align*} V = \pi \int_0^4{ \left( 3 - \sqrt{y} \right) ^2 - 1^2 \, dy} \end{align*}$
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  3. #3
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    Re: Volume of a revolving solid...

    Lets look at this step by step. Drawing a line from the given curve to the axis of rotation gives a radius of the circle the point on the curve follows as the curve is rotated around the axis. The area of the disk formed is \pi r^2. Taking the "thickness" of that line to be the differential "du" (u is perpendicular to the radius so along the axis of rotation) the volume is \pi r^2du. The volume of the entire figure is \int \pi r^2 du.

    Suppose R is the region bounded by y = x2, x = 2, and y = 0

    1) Find the volume of a solid that is generated by revolving R about the x-axis. - I believe I did this one correctly, but I want to be sure. I got 32pi/5.
    The distance from the point (x, y) to (x, 0) (the x-axis) is y= x^2 so r= x^2, r= (x^2)^2= x^4 and the integral is \pi \int_0^2 x^4 dx. Yes, that is \frac{32}{5}\pu.

    2) (same) ... revolving R about the y-axis. - I feel the same about this one. I got 8pi.
    Since the region does not extend all the way to the y-axis, this really a "washer". But you can do it as two disks. Calculate the volume to the outer boundary, x= 2, calculate the volume to the inner boundary,  x= \sqrt{y}, and subtract.
    The radius of the "outer boundary" is the constant, 2, so rotation forms a cylinder with radius 2 and height 2^2= 4 so volume \pi r^2h= \pi (4)(4)= 16\pi. The radius of the "inner boundary" is \sqrt{y} so each disk has area \pi r^2= \pi y. Taking the "thickness" to be dy, the volume of the outer portion is \pi \int_0^4 ydy= 8\pi. The difference is 16\pi- 8\pi= 8\pi, as you say.

    3) ... revolving R about y = -3.
    The distance from y= x^2 to y= -3 is r= x^2- (-3)= x^2+ 3 so the area of each "disk" is \pi(x^2+ 3)^2 and, taking "dx" as the thickness the volume of each disk is \pi(x^2+ 3)^2 dx. The total volume is \pi \int_0^2 (x^2+ 3)^2 dx.

    4) ... revolving R about x = 3.
    The distance from (x, y) to (3, y) is r= 3- x= 3- \sqrt{y} (x= 3 is to the right of the region). The area of a disk is \pi r^2= \pi(3- \sqrt{y})^2= \pi(y- 6\sqrt{y}+ 9). Taking the thickness of a disk to be dy, the volume of each disk is \pi (y- 6\sqrt{y}+ 9)dy and the total volume is \pi\int_0^4(y- 6y^{1/2}+ 9)dy.
    Last edited by HallsofIvy; April 29th 2014 at 10:02 AM.
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  4. #4
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    Re: Volume of a revolving solid...

    Thanks so much to both of you - ProveIt for the quick response, and HallsofIvy for going in-depth.
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