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nicoletugay
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Tyro
Asked: 2021-04-07T06:37:26+00:00 In: ,

Calculus Problem applying Cylindrical Shell Method

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Find the volume of the solid generated when the indicated plane region is revolved about
the axis of revolution. The region bounded by y = x
2 − 2x + 1, y = 7 − x about y = −1.

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  2. acedstud
    This answer was edited.

    The sketch of the solid formed by the revolving the region bounded by y=x^2-2x+1 and y=7-x about y=-1 is shown in the attached figure.

    From the figure, if we slice out tiny pieces of the solid, we will have cylinders with a height of dx, and the base area of \pi r^2, where r is the radius of the outer shell minus the radius of the inner shell. Notice that the outer shell is made by the line y=7-x and because the axis of revolution is y=-1, the radius of the outer shell (distance from the axis of revolution) is given by y=7-x+1=8-x. Similarly, the inner shell is made by the curve y=x^2-2x+1 and because the axis of revolution is y=-1, the radius of the inner shell (distance from the axis of revolution) is given by y=x^2-2x+1+1=x^2-2x+2.

    Thus, the base area of each slice of cylinder is given by \pi(8-x)^2-\pi(x^2-2x+2)^2=\pi[(64-16x+x^2)-(x^4-4x^3+6x^2-4x+1)].

    =\pi(64-16x+x^2-x^4+4x^3-6x^2+4x-1)=\pi(63-12x-5x^2+4x^3-x^4).

    Thus, the volume of the solid formed by revolving the region bounded by y=x^2-2x+1 and y=7-x about y=-1 is given by \pi\int_{-2}^3{63-12x-5x^2+4x^3-x^4}dx=\pi\left[63x-6x^2-\dfrac{5}{3}x^3+x^4-\dfrac{1}{5}x^5\right]_{-2}^3

    =\pi[122.4-(-114.3)]=\pi(122.4+114.3)=236.7\pi\approx743.5 cubic units.

    Attachment

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