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Math Help - Let R be the region ... find volume when R is revolved around x-axis

  1. #1
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    Let R be the region ... find volume when R is revolved around x-axis

    let R be the region in the first quadrant between the graphs of y=e^{-x}, y= sinx and the y-axis. The volume of the solid that results when R is revolved about the x-axis is

    (A) -0.888
    (B) -0.869
    (C) 0.277
    (D) 0.869
    (E) 0.888

    first thing i did was graph it on calculator then find where the graphs intersect but i forgot what to do from there
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  2. #2
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    First of all, volume can't be negative so A and B are out.
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  3. #3
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    Do you know the equation of finding the volume?
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    Quote Originally Posted by dwsmith View Post
    Do you know the equation of finding the volume?
    \pi \int_a^b (OR)^2 - (IR)^2 dx

    where OR= the outer radius and IR= the inner radius

    here's my attempt at the problem

    a≈0.588532
    b≈ 3.096364

    \pi \int_a^b (sinx)^2 - (e^{-x})^2 dx

    using calculator to find the definite integral i get

    \pi \int_a^b (sinx)^2 - (e^{-x})^2 dx ≈1.354
    problem is that is not one of the options which means i did something wrong.
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  5. #5
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    Just to verify the answer is E right?
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    Just to verify the answer is E right?
    yes the answer is E but i don't know how to get the answer
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  7. #7
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    I think you are misinterpreting the question. The region you are looking for starts at y=0 to the intersection.

    So your bounds of integration on the x axis are 0 to .588.

    I think you may have done the region further to right.
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    I think you are misinterpreting the question. The region you are looking for starts at y=0 to the intersection.

    So your bounds of integration on the x axis are 0 to .588.

    I think you may have done the region further to right.
    thanks
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