relskid
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Homework Statement
find the volume of the solid generated by rotating the circle (x-10)^2 + y^2 = 36 about the y-axis
Homework Equations
disk method: \pi\int [R(x)]^2dx
shell method: 2\pi\int (x)(f(x))dx
The Attempt at a Solution
y = \sqrt{36-(x-10)^2}dx[\tex]<br /> <br /> \\\pi\int [(\sqrt{36-(x-10)^2})]^2dx
\pi\int (36-(x-10)^2)dx
\pi\int (36-(x^2-20x-100))dx
\pi\int (-x^2+20x-64)dx
\pi [(\frac{-x^3}{3}+10x^2-64x)]
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:
Homework Statement
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.
Homework Equations
F =\int (p)(h(y))(L(y))dy
p=rho (density)
The Attempt at a Solution
x^2 + y^2 = 2^2
x^2 = 4 - y^2
x = \sqrt{4 - y^2}
note: integration from -1 to 0
42\int(-y)\sqrt{4 - y^2}dy
after that, i don't really know what to do. this is the part that I'm especially not sure about:
-42\int(y)\sqrt{4 - y^2}dy
u=4-y^2
du=-2ydy
-\frac{1}{2}du=ydy
-42(-\frac{1}{2})\int\sqrt{u}du
21\int\sqrt{u}du
21[\frac{u^\frac{3}{2}}{3/2}]
thanks to any who can help.