Question #261657

Find the volume generated by revolving the given region about the given axis.

(a) The region bounded by y = x4 , x = 1 and y = 0 about Y axis.

(b) The triangle with vertices (1, 1),(1, 2)(2, 2) about X axis.

(c) The region in the first quadrant bounded by x = y − y3 , x = 1 and y = 1 about X axis. 


1
Expert's answer
2021-11-08T19:50:26-0500

(a)


A1(y)=πx2=πyA_1(y)=\pi x^2=\pi\sqrt{y}

A2(y)=π(x)2=πA_2(y)=\pi(x)^2=\pi

V=01(ππy)dy=π[y23y3/2]10V=\displaystyle\int_{0}^{1}(\pi-\pi\sqrt{y})dy=\pi[y-\dfrac{2}{3}y^{3/2}]\begin{matrix} 1 \\ 0 \end{matrix}

=π3(cubic units)=\dfrac{\pi}{3} (cubic\ units)

(b)


V=12(π(2)2πx2)dy=π[4x13x3]21V=\displaystyle\int_{1}^{2}(\pi(2)^2-\pi x^2)dy=\pi[4x-\dfrac{1}{3}x^{3}]\begin{matrix} 2\\ 1 \end{matrix}

=π(883(413)=53(cubic units)=\pi(8-\dfrac{8}{3}-(4-\dfrac{1}{3})=\dfrac{5}{3}(cubic\ units)

(c)



V=012πy(1(yy3))dy=2π[y22y33+y44]10V=\displaystyle\int_{0}^{1}2\pi y(1-(y − y^3))dy=2\pi[\dfrac{y^2}{2}-\dfrac{y^3}{3}+\dfrac{y^4}{4}]\begin{matrix} 1\\ 0 \end{matrix}

=5π12(cubic units)=\dfrac{5\pi}{12}(cubic\ units)


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