Answer to Question #348512 in Calculus for Migz

Question #348512

The region in the first quadrant, which is bounded by the curve


x² = 4y, the line x = 4, is revolved about the line x = 4. Locate the


centroid of the resulting solid of revolution.

1
Expert's answer
2022-06-09T14:55:34-0400
x=4:42=4y=>y=4x=4: 4^2=4y=>y=4

x2=4y,x0=>:x=2y,y0x^2=4y, x\ge0=>: x=2\sqrt{y}, y\ge 0


V=π04(42(2y)2)dyV=\pi\displaystyle\int_{0}^{4}(4^2-(2\sqrt{y})^2) dy

=π[16y2y2]40=32π=\pi[16y-2y^2]\begin{matrix} 4 \\ 0 \end{matrix}=32\pi

In this case, the centroid lies on the axis x=4.x=4.

Vy=π04y(2y)dyV_y=\pi\displaystyle\int_{0}^{4}y(2\sqrt{y}) dy

=π[4y5/25]40=128π5=\pi[\dfrac{4y^{5/2}}{5}]\begin{matrix} 4 \\ 0 \end{matrix}=\dfrac{128\pi}{5}

yˉ=128π532π=0.8\bar{y}=\dfrac{\dfrac{128\pi}{5}}{32\pi}=0.8


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