Volume of solid and fluid force

[relskid]

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

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

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

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

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

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

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:

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

$$u=4-y^2$$

$$du=-2ydy$$

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

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

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

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

??

 

[mighty2000]

Thank you for posting your questions. I will do my best to help you find the solutions you are looking for.

For the first problem, you are on the right track but there are a few errors in your setup. First, in order to use the disk or shell method, you need to have the equation in terms of x or y. In this case, it would be easier to use y as the variable and rewrite the equation as y = sqrt(36 - (x-10)^2). This will make it easier to integrate. Also, when you square the function, you should get (36 - (x-10)^2), not (36 - (x-10)^2)^2. And finally, when you integrate, you need to use the correct limits of integration, which in this case would be from 4 to 16. So the correct setup would look like this:

V = pi * \int_4^16 (36 - (x-10)^2) dx

To solve this integral, you can expand the squared term and then use the power rule to integrate.

For the second problem, your setup is correct but there are a few errors in your integration. First, when you substitute u = 4-y^2, the limits of integration should also change. Since the original limits were from -1 to 0, the new limits would be from 4 to 1. Also, when you integrate, you should get 21 * (u^(3/2) / (3/2)), not (u^(3/2) / (3/2)). And finally, don't forget to substitute back in for u and use the correct limits of integration to get your final answer.

I hope this helps you solve these problems. Let me know if you have any other questions or if you need further clarification. Good luck with your studies!