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Math Help - Finding the area of a solid with Disc/Washer Method

  1. #1
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    Finding the area of a solid with Disc/Washer Method

    So, I'm quite rusty on Calculus and was wondering if anyone could offer some help-- I'd really appreciate it! I'm also new to this site so pardon if I don't know how to format things.


    1) I'm asked to find the volume of a solid by rotating the region bound by^3, y=8, and x=0 about x= -4 by using the disc/washer method.

    I'm unsure how to proceed because I don't know if I'm supposed to set them all up to solve for y or x because I have both y=8 and x=0. Since I'm rotating around the vertical line, x=-4, would I solve for y? And if so, how do I set up the integrals y=8 and x=0?


    2) I I'm asked to find the volume of an area bounded by y=5 and y=x+(4/x) that revolves about x= -1. When I set up the integral, which function goes first?

    I'd appreciate any assistance!
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  2. #2
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    Re: Finding the area of a solid with Disc/Washer Method

    In the first question, what is the first function bounding the region?
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    Re: Finding the area of a solid with Disc/Washer Method

    Sorry, I meant by y=x^3, not by^3
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    Re: Finding the area of a solid with Disc/Washer Method

    If you picture the area as a sum of strips, since you are rotating about a vertical line, the strips will be horizontal. So that means when you rotate them and get washers, the inner and outer radii will be horizontal lines and the heights will be small changes in y. So you need to write down these radii first in terms of x, THEN in terms of y, so that you can perform the integration with respect to y (as the small changes in height are small changes in y, namely dy).
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    Re: Finding the area of a solid with Disc/Washer Method

    Quote Originally Posted by Prove It View Post
    So you need to write down these radii first in terms of x, THEN in terms of y, so that you can perform the integration with respect to y (as the small changes in height are small changes in y, namely dy).
    So I solve for x (ex: y=x^3 becomes x=y^(1/3))? But then how does that change y=8? And what happens to x=0? Do those functions only provide the bounds 0 to 8 for which I will integrate the function y^(1/3) or are they also integrated?

    Thanks for the help!
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    Re: Finding the area of a solid with Disc/Washer Method

    Like I said, you have to picture each strip physically being rotated about the line x = -4. A diagram might help. From that line, the outside of the washer is given by the x-position of the cubic function. What distance is that from x = -4 (mind you, it will be variable)? That is your outer radius. The inside of the washer is given by the x-position of the line x = 0, a constant function, so it should be easy to determine the distance from x = -4. This is the inner radius.

    Do you understand how to evaluate the area of a circle? How to use that to evaluate the area of a washer? How to use that to evaluate the volume of a washer if its height is a small change in y (dy)?
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