Asked in calc area, wasn't answered

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The discussion focuses on solving two calculus problems involving volume and fluid force. For the first problem, the volume of the solid generated by rotating a circle about the y-axis can be approached using either the disk or shell method, with a recommendation to use the shell method for simplicity. The user expresses uncertainty about the limits of integration, which should be clarified by sketching the problem. In the second problem, the fluid force on a circular end of a horizontal cylindrical gasoline tank is calculated, with guidance provided on setting up the integral correctly. Overall, the conversation emphasizes the importance of choosing the appropriate method and understanding the integration limits.
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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.
 
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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.
 
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