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Math Help - Finding the Area/Volume of a solid using the shell/disk/washer method..

  1. #1
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    Finding the Area/Volume of a solid using the shell/disk/washer method..

    Ive been working on this problem for a while, and I can't get very far with it. I don't really know what to do.

    Let R be the region bounded by the graphs of y=√x, y=e^-x and the y axis.

    a) Find the area of R

    b) Find the volume of the solid generated when R is revolved about the horizontal line y=-1

    c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a semi circle whose diameter runs from the graph y=√x to the graph of y=e^-x. Find the volume of the solid.

    For A, I think Im supposed to used the formula A=bh. And becasue of the way the graph looks I think its supposed to be A=(√x)(e^-x). But I don't really know where to go from there.

    For B, I think I'm supposed to use the formula v=∫2πrh dx

    And for C, I don't really have any idea. From looking at my notes, I have the area formula of a semi-circle, which is A=πd^2/8...
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  2. #2
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    Part a is just asking for the area between the two curves. Basic subtraction. Take the area of the upper curve and subtract out the area of the lower curve. Your intervals will simply be the points where √x=e^-x.

    For part b, yes, you can use shell method. Your radius will be your distance from the x-axis (which will be y), and your height will be the distance between the two points on either curve (again, basic subtraction. One function - the other = distance between).

    Part c I can kind of conceptualize, but i'm failing to see how the region R is in tune with each cross-section being a semi-circle. Or how that semi-circle has a diameter attributed to it. But then, i've never been good with word-games.
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  3. #3
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    I kind of worked them out the way I thought I should...

    A) I have -e^-x - 2/3x^3/2 from 0,1. I plugged in, subtracted and got an answer of .310688483.

    For B I used the shell method, but I got a negitive answer.. I have -.693680127 π.

    and for C, I integrated the semi-circle formula with the diameter as (√x+e^-x). And I got a really small number.

    I think I did them all pretty wrong. Thanks for your help though. I know its probably really hard not being able to see the work I did. But I kind of understand more of what I'm doing.
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