# Thread: Parametric Solid

1. ## 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

$
r(t) = (\frac{1}
{4}e^{4t} - t){\mathbf{i}} + (e^{2t} ){\mathbf{j}}
$

Here is what I did...

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

Is this right? I have no answer key so I have no way to check. Thanks

2. Originally Posted by billa
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

$
r(t) = (\frac{1}
{4}e^{4t} - t){\mathbf{i}} + (e^{2t} ){\mathbf{j}}
$

Here is what I did...

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

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)
$

$\frac{\pi}{8}\left(e^8 - 1\right)^2$

3. arrrrgg

ty