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Math Help - Parametric Solid

  1. #1
    Member billa's Avatar
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    Parametric Solid

    Find the volume of the solid created by revolving the curve C about the x-axis. C is defined by the curve r(t)=[(.25)e^(4t)-t]*i + e^(2t)*j where t is between 0 and 2 including 0 and 2

    Here is r again

    <br />
r(t) = (\frac{1}<br />
{4}e^{4t}  - t){\mathbf{i}} + (e^{2t} ){\mathbf{j}}<br />

    Here is what I did...

    <br />
V = \int\limits_0^2 {\pi y^2 dx = } \int\limits_0^2 {\pi y^2 \frac{{dx}}<br />
{{dt}}dt = } \int\limits_0^2 {\pi e^{4t} (e^{4t}  - 1)dt = } \pi e^{16}  - \pi e^8 <br />


    Is this right? I have no answer key so I have no way to check. Thanks
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by billa View Post
    Find the volume of the solid created by revolving the curve C about the x-axis. C is defined by the curve r(t)=[(.25)e^(4t)-t]*i + e^(2t)*j where t is between 0 and 2 including 0 and 2

    Here is r again

    <br />
r(t) = (\frac{1}<br />
{4}e^{4t}  - t){\mathbf{i}} + (e^{2t} ){\mathbf{j}}<br />

    Here is what I did...

    <br />
V = \int\limits_0^2 {\pi y^2 dx = } \int\limits_0^2 {\pi y^2 \frac{{dx}}<br />
{{dt}}dt = } \int\limits_0^2 {\pi e^{4t} (e^{4t}  - 1)dt = } \pi e^{16}  - \pi e^8 <br />


    Is this right? I have no answer key so I have no way to check. Thanks
    \pi \int_0^2 e^{4t} (e^{4t}  - 1) \, dt

    \pi \int_0^2 e^{8t} - e^{4t} \, dt

    \pi \left[\frac{e^{8t}}{8} - \frac{e^{4t}}{4}\right]_0^2

    \pi \left[\left(\frac{e^{16}}{8} - \frac{e^{8}}{4}\right)-\left(\frac{1}{8} - \frac{1}{4}\right)\right]

    \frac{\pi}{8}\left(e^{16} - 2e^8 + 1\right)<br />

    \frac{\pi}{8}\left(e^8 - 1\right)^2
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  3. #3
    Member billa's Avatar
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    arrrrgg


    ty
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