(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

find the volume of the solid generated by rotating the circle (x-10)^2 + y^2 = 36 about the y-axis

2. Relevant equations

disk method: [tex]\pi\int [R(x)]^2dx[/tex]

shell method: [tex]2\pi\int (x)(f(x))dx[/tex]

3. The attempt at a solution

[tex]y = \sqrt{36-(x-10)^2}dx[\tex]

\\\pi\int [(\sqrt{36-(x-10)^2})]^2dx[/tex]

[tex]\pi\int (36-(x-10)^2)dx[/tex]

[tex]\pi\int (36-(x^2-20x-100))dx[/tex]

[tex]\pi\int (-x^2+20x-64)dx[/tex]

[tex]\pi [(\frac{-x^3}{3}+10x^2-64x)][/tex]

ok, as you may have noticed, the integral isn't definite. that's because i don't know whether it should be from 4 to 16, or -6 to 6. also, if i did the entire problem wrong, that'd be nice to know, too. :P

next problem:

1. The problem statement, all variables and given/known data

a cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. find the fluid force on a circular end of the tank if the tank is half full, assuming that the diameter is 2 feet and the gasoline weighs 42 pounds per cubic foot.

2. Relevant equations

[tex]F =\int (p)(h(y))(L(y))dy[/tex]

p=rho (density)

3. The attempt at a solution

[tex]x^2 + y^2 = 2^2[/tex]

[tex]x^2 = 4 - y^2 [/tex]

[tex]x = \sqrt{4 - y^2} [/tex]

note: integration from -1 to 0

[tex]42\int(-y)\sqrt{4 - y^2}dy[/tex]

after that, i don't really know what to do. this is the part that i'm especially not sure about:

[tex]-42\int(y)\sqrt{4 - y^2}dy[/tex]

[tex]u=4-y^2[/tex]

[tex]du=-2ydy[/tex]

[tex]-\frac{1}{2}du=ydy[/tex]

[tex]-42(-\frac{1}{2})\int\sqrt{u}du[/tex]

[tex]21\int\sqrt{u}du[/tex]

[tex]21[\frac{u^\frac{3}{2}}{3/2}][/tex]

thanks to any who can help.

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# Homework Help: Asked in calc area, wasn't answered

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