ANSWER The centroid is ( 8 3 , 0 , 0 ) (\frac{8}{3},0,0) ( 3 8 , 0 , 0 )
EXPLANATION
If x 0 x_{0} x 0 is fixed, then when rotating the segment { ( x 0 , y , 0 ) : 0 ≤ y ≤ 2 x 0 } \left \{\left ( x_{0},y,0 \right ) :0\leq y\leq 2\sqrt{x_{0}} \right \} { ( x 0 , y , 0 ) : 0 ≤ y ≤ 2 x 0 } around x-axis, we get a circle : y 2 + z 2 = 4 x 0 y^{2}+z^{2}=4x_{0} y 2 + z 2 = 4 x 0 . So, the resulting solid of revolution is
R = { ( x , y , z ) : 0 ≤ x ≤ 4 , 0 ≤ y 2 + z 2 ≤ 4 x } R=\left \{ \left ( x,y,z \right ): 0\leq x\leq 4,\, 0\leq y^{2} +z^{2}\leq 4x\right \} R = { ( x , y , z ) : 0 ≤ x ≤ 4 , 0 ≤ y 2 + z 2 ≤ 4 x }
The centroid ( x C , y C , z C ) (x_{C},y_{C},z_{C}) ( x C , y C , z C ) is given by
x C = ∭ R x d V ∭ R d V , y C = ∭ R y d V ∭ R d V , z C = ∭ R z d V ∭ R d V 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} x C = ∭ R d V ∭ R x d V , y C = ∭ R d V ∭ R y d V , z C = ∭ R d V ∭ R z d V
∭ R d V = ∫ 0 4 d x ∬ S x d y d z \iiint_{R}dV=\int_{0}^{4}dx\iint_{S_{x}} dydz ∭ R d V = ∫ 0 4 d x ∬ S x d y d z ,
where S x S_{x} S x is a circle whose radius is 4 x 4x 4 x :
S x = { ( y , z ) : y 2 + z 2 ≤ 4 x } S_{x }= \left \{ \left ( y,z \right ) :y^{2}+z^{2}\leq 4x\right \} S x = { ( y , z ) : y 2 + z 2 ≤ 4 x }
A r e a o f S x = ∬ S x d y d z = 4 π ⋅ x Area \, \, of\, \, S_{x}=\iint_{S_{x}} dydz=4\pi\cdot x A re a o f S x = ∬ S x d y d z = 4 π ⋅ x ,
so ∭ R d V = 4 π ∫ 0 4 x d x = 4 π [ x 2 2 ] 0 4 = 2 π ( 16 − 0 ) = 32 π \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 ∭ R d V = 4 π ∫ 0 4 x d x = 4 π [ 2 x 2 ] 0 4 = 2 π ( 16 − 0 ) = 32 π
∭ R x d V = ∫ 0 4 x d x ∬ S x d y d z = 4 π ∫ 0 4 x 2 = 4 π [ x 3 3 ] 0 4 = 256 π 3 ⇒ x C = 8 3 ≅ 2.6667 \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 ∭ R x d V = ∫ 0 4 x d x ∬ S x d y d z = 4 π ∫ 0 4 x 2 = 4 π [ 3 x 3 ] 0 4 = 3 256 π ⇒ x C = 3 8 ≅ 2.6667
Now, to calculate integrals over the region S x S_{x} S x we replace the Cartesian coordinates with polar coordinates : y = r cos t , z = r sin t y=r\cos t, z=r\sin t y = r cos t , z = r sin t ( 0 ≤ t ≤ 2 π , 0 ≤ r ≤ 2 x ) (0\leq t\leq 2\pi, 0\leq r\leq 2\sqrt{x}) ( 0 ≤ t ≤ 2 π , 0 ≤ r ≤ 2 x ) . Therefore
∭ R y d V = ∫ 0 4 d x ∬ S x y d y d z = ∫ 0 4 [ ∫ 0 2 π ∫ 0 2 x r 2 cos t d r d t ] d x = = ( ∫ 0 2 π cos t d t ) ( ∫ 0 4 ∫ 0 2 x r 2 d r d x ) = 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 , ∭ R y d V = ∫ 0 4 d x ∬ S x y d y d z = ∫ 0 4 [ ∫ 0 2 π ∫ 0 2 x r 2 cos t d r d t ] d x = = ( ∫ 0 2 π cos t d t ) ( ∫ 0 4 ∫ 0 2 x r 2 d r d x ) = 0 ,
because ∫ 0 2 π cos t d t = 0. \int_{0}^{2\pi}\cos tdt =0. ∫ 0 2 π cos t d t = 0.
∭ R z d V = ∫ 0 4 d x ∬ S x z d y d z = ∫ 0 4 [ ∫ 0 2 π ∫ 0 2 x r 2 sin t d r d t ] d x = = ( ∫ 0 2 π sin d t ) ( ∫ 0 4 ∫ 0 2 x r 2 d r d x ) = 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 , ∭ R z d V = ∫ 0 4 d x ∬ S x z d y d z = ∫ 0 4 [ ∫ 0 2 π ∫ 0 2 x r 2 sin t d r d t ] d x = = ( ∫ 0 2 π sin d t ) ( ∫ 0 4 ∫ 0 2 x r 2 d r d x ) = 0 ,
since ∫ 0 2 π sin t d t = 0 \int_{0}^{2\pi} \sin tdt =0 ∫ 0 2 π sin t d t = 0 . Hence y C = 0 , z C = 0 y_{C}=0, z_{C}=0 y C = 0 , z C = 0
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