Answer to Question #348512 in Calculus for Migz

Question

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.

Answer 1

$$x=4: 4^2=4y=>y=4$$

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

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

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

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

$$V_y=\pi\displaystyle\int_{0}^{4}y(2\sqrt{y}) dy$$

$$=\pi[\dfrac{4y^{5/2}}{5}]\begin{matrix} 4 \\ 0 \end{matrix}=\dfrac{128\pi}{5}$$

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