Question #348511

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.

1
Expert's answer
2022-06-09T14:02:10-0400

ANSWER The centroid is (83,0,0)(\frac{8}{3},0,0)

EXPLANATION

If x0x_{0} is fixed, then when rotating the segment {(x0,y,0):0y2x0}\left \{\left ( x_{0},y,0 \right ) :0\leq y\leq 2\sqrt{x_{0}} \right \} around x-axis, we get a circle : y2+z2=4x0y^{2}+z^{2}=4x_{0} . So, the resulting solid of revolution is

R={(x,y,z):0x4,0y2+z24x}R=\left \{ \left ( x,y,z \right ): 0\leq x\leq 4,\, 0\leq y^{2} +z^{2}\leq 4x\right \}



The centroid (xC,yC,zC)(x_{C},y_{C},z_{C}) is given by

xC=RxdVRdV,yC=RydVRdV,zC=RzdVRdVx_{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}

RdV=04dxSxdydz\iiint_{R}dV=\int_{0}^{4}dx\iint_{S_{x}} dydz ,

where SxS_{x} is a circle whose radius is 4x4x :

Sx={(y,z):y2+z24x}S_{x }= \left \{ \left ( y,z \right ) :y^{2}+z^{2}\leq 4x\right \}

AreaofSx=Sxdydz=4πxArea \, \, of\, \, S_{x}=\iint_{S_{x}} dydz=4\pi\cdot x,

so RdV=4π04xdx=4π[x22]04=2π(160)=32π\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

RxdV=04xdxSxdydz=4π04x2=4π[x33]04=256π3xC=832.6667\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 SxS_{x} we replace the Cartesian coordinates with polar coordinates : y=rcost,z=rsinty=r\cos t, z=r\sin t (0t2π,0r2x)(0\leq t\leq 2\pi, 0\leq r\leq 2\sqrt{x}) . Therefore

RydV=04dxSxydydz=04[02π02xr2costdrdt]dx==(02πcostdt)(0402xr2drdx)=0,\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 02πcostdt=0.\int_{0}^{2\pi}\cos tdt =0.

RzdV=04dxSxzdydz=04[02π02xr2sintdrdt]dx==(02πsindt)(0402xr2drdx)=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 02πsintdt=0\int_{0}^{2\pi} \sin tdt =0 . Hence yC=0,zC=0y_{C}=0, z_{C}=0


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