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Math Help - Find the volume of a solid using both the disk/washer and shell method.

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    Find the volume of a solid using both the disk/washer and shell method.

    Let R be the region bounded by the following curves. Let S be the solid generated  when R is revolved about the given axis. If possible, find the volume of S by both the disk/washer and shell methods. Check that your results agree.

    I have attached an image with the functions and the correct answers.


    I have tried integrating this both as f(x) and f(y), but both lead me to have to take anti-derivatives that seem impossible with the calculus i have learned so far (u-substitution would not work).

    I tried to type some of my work here, but when it posted it came out in a bunch of random letters and numbers and such.

    Thank you for the help!
    Attached Thumbnails Attached Thumbnails Find the volume of a solid using both the disk/washer and shell method.-math.jpg  
    Last edited by tanagholat; June 23rd 2012 at 10:55 AM.
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    Re: Find the volume of a solid using both the disk/washer and shell method.

    I don't see why it would give difficult integrals. Rotating around the x-axis gives disks with radii y and so area \pi y^2. The square root, after squaring, will be gone so you just have integrals of logarithms. What is the integral of \int ln(ax) dx?
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    Re: Find the volume of a solid using both the disk/washer and shell method.

    Quote Originally Posted by tanagholat View Post
    I have tried integrating this both as f(x) and f(y), but both lead me to have to take anti-derivatives that seem impossible with the calculus i have learned so far (u-substitution would not work).
    For integrals involving logarithms, it's often useful to integrate by parts. For example, to integrate \int\ln x^n\,dx,\ n\geq1 let

    u = \ln x^n\Rightarrow du=\frac{nx^{n-1}}{x^n}\,dx

    \Rightarrow du=\frac nx\,dx,

    and let dv = dx\Rightarrow v = x, which gives

    \int\ln x^n\,dx = \int u\,dv

    =uv - \int v\,du

    =x\ln x^n - \int n\,dx

    =x\left(\ln x^n - 1\right) + C

    Now, to the problem at hand.



    Washer Method:

    First we determine where y=1 intersects the other two curves:

    \sqrt{\ln\frac{x^2}{36}} = 1\Rightarrow x = 6\sqrt e (ignoring the negative root)

    \sqrt{\ln\frac x6} = 1\Rightarrow x = 6e

    So,

    V = \pi\int_6^{6\sqrt e}\left[\left(\sqrt{\ln\frac{x^2}{36}}\right)^2 - \left(\sqrt{\ln\frac x6}\right)^2\right]\,dx + \pi\int_{6\sqrt e}^{6e}\left[1^2 - \left(\sqrt{\ln\frac x6}\right)^2\right]\,dx

    = \pi\int_6^{6\sqrt e}\left(\ln\frac{x^2}{36} - \ln\frac x6\right)\,dx + \pi\int_{6\sqrt e}^{6e}\left(1 - \ln\frac x6}\right)\,dx

    = \pi\left[x\left(\ln\frac{x^2}{36} - 2\right) - x\left(\ln\frac x6 - 1\right)\right]_6^{6\sqrt e} + \pi\left[x - x\left(\ln\frac x6 - 1\right)\right]_{6\sqrt e}^{6e}

    =\pi\left[\left(6\sqrt e\right)(-1) - \left(6\sqrt e\right)\left(-\frac12\right) - 6(-2) + 6(-1) + 6e - 6e(0) - 6\sqrt e + \left(6\sqrt e\right)\left(-\frac12\right)\right]

    =6\pi\left(e - 2\sqrt e + 1\right)

    =6\pi\left(\sqrt e - 1\right)^2



    Shell Method:

    y = \sqrt{\ln\frac x6}\Rightarrow x = 6e^{y^2}

    y = \sqrt{\ln\frac{x^2}{36}}\Rightarrow x = 6e^{y^2/2}

    V = 2\pi\int_0^1y\left(6e^{y^2} - 6e^{y^2/2}\right)\,dy

    This integral is relatively straightforward. I leave it to you to show that it produces the same result.
    Thanks from tanagholat
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    Re: Find the volume of a solid using both the disk/washer and shell method.

    Thank you very much to the both of you.

    So we haven't learned how to do integration by parts with logarithms and such, so i guess that's where I got stuck, but other than that it's nice to know that my thought process was all correct. I did some research on how you got that and it makes some sense now. (yay)

    Much appreciation to the both of you. I'm now a bit more confident with knowledge of this material.
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