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anonymous
 4 years ago
Find the volume of the solid generated by revolving the region bounded by the xaxis 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.
anonymous
 4 years ago
Find the volume of the solid generated by revolving the region bounded by the xaxis 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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dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1For 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 crosssection is a ring with outer radius of 3 inner radius of 3f(x) = 3xsinx \[V =\pi \int\limits_{0}^{\pi}R^{2} r^{2} \] \[=\pi \int\limits_{0}^{\pi}3^{2}(3x \sin x)^{2} dx\] \[=\pi \int\limits_{0}^{\pi}6x \sin xx^{2}\sin^{2} x dx\] 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 = 2pix height is just the y_value or f(x) \[V = 2\pi \int\limits_{0}^{\pi}(2\pix)(x \sin x) dx\] part c) dw:1328794750169:dw each circular crosssection 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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