Answer to Question #193255 in Calculus for April

Question

Find the centroid of the solid generated if the region bounded by y = 2- x^2, x=0, and y=0 is revolved about the y-axis.

Answer 1

Answer:-

$y=2-x^2$ , x=0 , y=0

Volume =$\int_0^2x^2dy=\pi\int_0^2(2-y)dy$

=$\pi[2y-{1\over 2}y^2]_0^2$

=$\pi[(2(2)-{1\over 2}(2)^2)-(2(0)-{1\over 2}(0)^2)]=2\pi$

$\therefore$ Volume = $2\pi$

find the coordinate for the centroid

$2-x^2=0 \implies x=$$2^{1\over 2}$ $\implies x$ =1.4142135623731 or x=-1.4142135623731 use positive value since we are given x=0 and y=0

y=f(x)=$2-x^2$

x coordinate=${1\over Volume}\int_0^{1.4142135623731} xf(x)dx$ = ${1\over Area}\int_0^{1.4142135623731}$ x(2-$x^2$ )dx

$=\int_0^{1.4142135623731}(2x-x^3)dx$

=${1\over 2\pi}$ $([{2 x^2\over 2}-{x^4\over 4}]_0^{1.4142135623731})$

=${1\over 2\pi}$ $(({2( 1.4142135623731)^2\over 2}-{1.4142135623731^4\over 4})-({2(0)^2\over 2}-{0^4\over 4}))$

=0.26

y=$2-0^2$ =2

x=f(y)=$(2-y)^{1\over 2}$

y coordinate=${1\over Volume}\int_0^2 yf(y)dy$ =${1\over 2\pi }$ $\int_0^2$ $y(2-y)^{1\over 2}dy$

let u=2-y $\implies$ du=-dy $\implies$ dy=-du

u=2-y $\implies$ y=2-u

when y=0 , u=2-0=2 and when y=2 ,u=2-2=0

$\int_0^2$ $y(2-y)^{1\over 2}dy$ = $-\int_2^0$ $(2-u)u^{1\over 2}du=-\int_2^0(2u^{1\over 2}-u^{3\over 2})du$

$-\int_2^0(2u^{1\over 2}-u^{3\over 2})du$ =$-[{4\over 3}u^{3\over 2}-{2\over 5}u^{5\over 2}]_2^0$

=$-[({4\over 3}(0)^{3\over 2}-{2\over 5}(0)^{5\over 2})-({4\over 3}(2)^{3\over 2}-{2\over 5}(2)^{5\over 2})]$

=-(-1.5084944665313)

=1.5084944665313

$\implies y coordinate= {1\over 2\pi}(1.5084944665313)$

=0.24

$\therefore$ the centroid is (0.26 , 0.24)