# Volume of solid and fluid force

1. Oct 31, 2007

### relskid

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: $$\pi\int [R(x)]^2dx$$

shell method: $$2\pi\int (x)(f(x))dx$$

3. The attempt at a solution

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

$$\pi\int (36-(x-10)^2)dx$$

$$\pi\int (36-(x^2-20x-100))dx$$

$$\pi\int (-x^2+20x-64)dx$$

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

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:
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

$$F =\int (p)(h(y))(L(y))dy$$

p=rho (density)

3. 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:

$$-42\int(y)\sqrt{4 - y^2}dy$$

$$u=4-y^2$$

$$du=-2ydy$$

$$-\frac{1}{2}du=ydy$$

$$-42(-\frac{1}{2})\int\sqrt{u}du$$

$$21\int\sqrt{u}du$$

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

??