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anonymous

  • 4 years ago

Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve xsinx , about 0<x<pi a) the y = 3 b) the line x = 2pi c) the y =-2 d) the line x = -2pi Please use the shell 2pirh formula. Steps please, so I can understand the r. I keep getting the r wrong. Thank you.

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  1. dumbcow
    • 4 years ago
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    For parts a) and c) i will use the disc method because to use shell method when revolving around horizontal axis means getting function in terms of y. y = xsin(x) --> x = f(y) = ? see what i mean part a) |dw:1328793504021:dw| each circular cross-section is a ring with outer radius of 3 inner radius of 3-f(x) = 3-xsinx \[V =\pi \int\limits_{0}^{\pi}R^{2} -r^{2} \] \[=\pi \int\limits_{0}^{\pi}3^{2}-(3-x \sin x)^{2} dx\] \[=\pi \int\limits_{0}^{\pi}6x \sin x-x^{2}\sin^{2} x dx\] part b) |dw:1328794143959:dw| the radius is distance from line of revolution as x goes from 0 to pi, radius goes from 2pi to pi --> r = 2pi-x height is just the y_value or f(x) \[V = 2\pi \int\limits_{0}^{\pi}(2\pi-x)(x \sin x) dx\] part c) |dw:1328794750169:dw| each circular cross-section is a ring with outer radius of 2+f(x) = 2+xsinx inner radius of 2 \[V =\pi \int\limits_{0}^{\pi}R^{2} -r^{2} \] \[=\pi \int\limits_{0}^{\pi}(2+x sin x)^{2}-2^{2} dx\] \[=\pi \int\limits_{0}^{\pi}4x \sin x+x^{2}\sin^{2} x dx\] part d) |dw:1328795102992:dw| the radius is distance from line of revolution (x=-2pi) as x goes from 0 to pi, radius goes from 2pi to 3pi --> r = 2pi+x height is just the y_value or f(x) \[V = 2\pi \int\limits_{0}^{\pi}(2\pi+x)(x \sin x) dx\]

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