- #1
clairez93
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Homework Statement
1. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.
[tex] y = \sqrt{x}, y = 0, x = 4[/tex]
the line x = 62. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y=4.
[tex] y = \frac{1}{x}, y = 0, x = 1, x = 4[/tex]
Homework Equations
The Attempt at a Solution
1.
[tex] V = \pi \int^{\sqrt{6}}_{0} ((6-y^{2})^{2} - (6-4)^{2}) dy[/tex]
[tex] = \pi \int^{\sqrt{6}}_{0} (36 - 12y^{2} + y^{4} -4) dy [/tex]
[tex] = \pi \int^{\sqrt{6}}_{0} (32 - 12y^{2} + y^{4}) dy [/tex]
[tex] = \pi [32y - 4y^{3} + \frac{y^{5}}{5}]^{\sqrt{6}}_{0} [/tex]
[tex] = \frac{76\sqrt{6}\pi}{5}[/tex]
I am relatively sure that I integrated the expression I set up correctly, because I checked it with my ti-89. That said, that means the original integral I set up is incorrect. I'm not sure why.
2.
[tex] V = \pi \int^{4}_{1} (4^{2} - \frac{1}{x^{2}}) dx [/tex]
[tex]= \pi \int^{4}_{1} (16 - \frac{1}{x^{2}}) dx [/tex]
[tex]= \pi [16x + \frac{1}{x}]^{4}_{1} [/tex]
[tex]= \pi [(64 + \frac{1}{4)}) - (16+1)][/tex]
[tex] = \pi (\frac{257}{4} - 17)[/tex]
[tex] = \frac{189\pi}{4}[/tex]
Again, checked with my calculator and I evaluated integral correctly. Set up went wrong.