Answer to Question #317168 in Civil and Environmental Engineering for Rodz

Question #317168

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.


1
Expert's answer
2022-03-24T01:41:23-0400

"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)


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