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Math Help - Cylindrical Shell Method + other Volume question

  1. #1
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    Cylindrical Shell Method + other Volume question

    Having trouble with these two problems:

    1) Region bounded by y= sqrt[x], y = 2, x =0

    Find the area using the shell method by revolving the enclosed region around the following axis:
    a) x-axis
    b) y-axis
    c) x=4
    d) y=2


    2) Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve y=x sin x, 0≤x≤pi about,
    a) y -axis
    b) the line x= pi

    Thanks!!
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  2. #2
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    Quote Originally Posted by Jeffman50 View Post
    Having trouble with these two problems:

    1) Region bounded by y= sqrt[x], y = 2, x =0

    Find the area using the shell method by revolving the enclosed region around the following axis:
    a) x-axis
    A line parallel to the x-axis at distance y from the x- axis would go from (0, y) to (y^2, y) and, rotated around the x-axis would form a cylinder of height y^2 and radius y. A circle around the x-axis with radius y has circumference 2\pi y and so the cylinder has surface area 2\pi y(y^2)= 2\pi y^3. Taking dy as the thickness, the volume of the thin shell would be 2\pi y^3 dy and the entire volume, 2\pi\int_{y= 0}^2 y^3 dy.

    b) y-axis
    It would be easier to use the "disk" method but since you specifically ask about the shell method: Rotating around the y-axis, a line parallel to the y-axis would go from y= \sqrt{x} to y= 2, so have length 2- \sqrt{x} and have radius x. Its surface area is 2\pi x(2- \sqrt{x}. The volume of such a thin shell is [tex]2\pi x(2- \sqrt{x})dx and the entired area is 2\pi\int_0^4 x(2- \sqrt{x})dx= 2\pi\int_0^4 (2x- x^{3/2})dx.

    c) x=4
    Instead of the radius being "x" as in (b), the radius would be "4- x".

    d) y=2
    Instead of the radius being "y" as in (a), the radius would be "2- y".


    2) Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve y=x sin x, 0≤x≤pi about,
    a) y -axis
    Using shells, each shell would have height xsin(x)- 0= xsin(x) and the radius would b x. The volume is 2\pi\int_0^\pi x(x sin(x)dx= 2\pi\int_0^\pi x^2 sin(x) dx.

    b) the line x= pi
    Same as a except the radius is \pi- x.

    Thanks!![/QUOTE]
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