Question

Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve xsinx , about 0 14 years ago

Answer 1

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)

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)

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)

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)

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)
\[V = 2\pi \int\limits_{0}^{\pi}(2\pi+x)(x \sin x) dx\]