Answer to Question #348511 in Calculus for Migz

Question

The region in the first quadrant which is bounded by the curve y² =

4x, and the lines x = 4 and y = 0, is revolved about the x-axis.

Locate the centroid of the resulting solid of revolution.

Answer 1

ANSWER The centroid is $(\frac{8}{3},0,0)$

EXPLANATION

If $x_{0}$ is fixed, then when rotating the segment $\left \{\left ( x_{0},y,0 \right ) :0\leq y\leq 2\sqrt{x_{0}} \right \}$ around x-axis, we get a circle : $y^{2}+z^{2}=4x_{0}$ . So, the resulting solid of revolution is

$R=\left \{ \left ( x,y,z \right ): 0\leq x\leq 4,\, 0\leq y^{2} +z^{2}\leq 4x\right \}$

The centroid $(x_{C},y_{C},z_{C})$ is given by

$x_{C}=\frac{\iiint_{R}xdV} {\iiint_{R}dV} ,\, \,y_{C}=\frac{\iiint_{R}ydV} {\iiint_{R}dV} ,\, \,z_{C}=\frac{\iiint_{R}zdV} {\iiint_{R}dV}$

$\iiint_{R}dV=\int_{0}^{4}dx\iint_{S_{x}} dydz$ ,

where $S_{x}$ is a circle whose radius is $4x$ :

$S_{x }= \left \{ \left ( y,z \right ) :y^{2}+z^{2}\leq 4x\right \}$

$Area \, \, of\, \, S_{x}=\iint_{S_{x}} dydz=4\pi\cdot x$,

so $\iiint_{R}dV=4\pi\int_{0}^{4}xdx=4\pi\left [ \frac{x^{2}}{2} \right ]_{0}^{4}=2\pi\left (16-0 \right )=32\pi$

$\iiint_{R}xdV=\int_{0}^{4}xdx\iint_{S_{x}} dydz=4\pi\int_{0}^{4} x ^{2}=4\pi\left [ \frac{x^{3}}{3} \right ]_{0}^{4}=\frac{256\pi}{3} \Rightarrow x_{C}=\frac{8}{3} \cong 2.6667$

Now, to calculate integrals over the region $S_{x}$ we replace the Cartesian coordinates with polar coordinates : $y=r\cos t, z=r\sin t$ $(0\leq t\leq 2\pi, 0\leq r\leq 2\sqrt{x})$ . Therefore

$\iiint_{R}ydV=\int_{0}^{4} dx\iint_{S_{x}}y dydz= \int_{0}^{4} \left [ \int_{0}^{2\pi} \int_{0}^{2\sqrt{x}}r^{2}\cos tdrdt\right ]dx=\\=\left ( \int_{0}^{2\pi}\cos tdt \right )\left ( \int_{0}^{4}\int_{0}^{2\sqrt{x}}r^{2}drdx \right )=0 ,$

because $\int_{0}^{2\pi}\cos tdt =0.$

$\iiint_{R}zdV=\int_{0}^{4} dx\iint_{S_{x}}z dydz= \int_{0}^{4} \left [ \int_{0}^{2\pi} \int_{0}^{2\sqrt{x}}r^{2}\\ \sin tdrdt\right ]dx=\\=\left ( \int_{0}^{2\pi}\\ \sin dt \right )\left ( \int_{0}^{4}\int_{0}^{2\sqrt{x}}r^{2}drdx \right )=0 ,$

since $\int_{0}^{2\pi} \sin tdt =0$ . Hence $y_{C}=0, z_{C}=0$