Asked in calc area, wasn't answered

[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:

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

$$u=4-y^2$$

$$du=-2ydy$$

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

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

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

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


thanks to any who can help.

 

[Midy1420]

Problem #1: from what you have you are using the disk method, which means you would have to switch variables (solve for x) and then it becomes kind of messy. Instead, use Shell method and use symmetry of the problem or you would need to do 2 integrals. That way you can use the way you did in terms of y = Sqrt(36-(x-10)^2). You'll see the limits right away when you draw it out. Hope that helps...hope I'm right as well.

If the representative rectangle is perpendicular to the axis of revolution, use the disk method.
If the representative rectangle is parallel to the axis of revolution, use the shell method.

 

[ogb p]

I would like to provide some guidance and tips for solving these problems.

For the first problem, you are on the right track with using the disk method for finding the volume of the solid generated by rotating the circle about the y-axis. However, your integration limits should be from -6 to 6, since the circle has a radius of 6. Also, when you are squaring the expression for y, you should have y^2 = 36 - (x-10)^2, not y^2 = 36 - (x-10)^2dx. The dx should not be included in the squared expression.

Once you have corrected these errors, you can continue with the integration to get the correct answer.

For the second problem, you are correct in setting up the equation for fluid force as F = ∫ρgh(y)L(y)dy, where ρ is the density of the fluid, g is the acceleration due to gravity, h(y) is the height of the fluid at a given point y, and L(y) is the length of the tank at that point.

To find the height of the fluid at any given point y, you can use the equation for a circle, x^2 + y^2 = 2^2, and solve for x. This will give you the expression for the radius of the circle at that point, which can then be used to find the height of the fluid.

Once you have the expression for h(y), you can continue with the integration to find the fluid force on the circular end of the tank. Remember to use the correct limits of integration and to substitute in the correct values for the density and length of the tank.

I hope this helps guide you in the right direction for solving these problems. Remember to always check your work and make sure your equations and units are consistent. Good luck!