Centroid of a Cylindrical Wedge

Calculus Level 4

A cylinder of radius 3 3 and length 10 10 is placed on the horizontal x y xy plane, with the center of it circular base at the origin. The plane z = 5 3 ( x + 3 ) z = \dfrac{5}{3} ( x + 3 ) cuts the cylinder in two halves. Take the lower half depicted in the figure above and find its centroid. If the centroid G = ( x 0 , y 0 , z 0 ) G = (x_0, y_0, z_0) , find 4 x 0 + 8 z 0 4 x_0 + 8 z_0 .


The answer is 28.

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1 solution

Hosam Hajjir
Dec 24, 2020

Taking a slice of the wedge at x = t x = t , where 3 t 3 -3 \le t \le 3 of thickness d t dt , then this slice will be a rectangle of dimensions d t × 2 9 t 2 × 5 3 ( t + 3 ) dt \times 2 \sqrt{9 - t^2} \times \dfrac{5}{3} (t + 3) , and the centroid of the slice is at ( t , 0 , 5 6 ( t + 3 ) ) ( t, 0, \dfrac{5}{6} (t + 3) )

The centroid G = ( 1 V ) t = 3 3 ( t , 0 , 5 6 ( t + 3 ) ) A ( t ) d t G = \displaystyle \left( \dfrac{1}{V} \right) \int_{t = -3 } ^{ 3 } ( t, 0, \dfrac{5}{6} (t + 3 ) ) A(t) dt

where A ( t ) A(t) is the area of the slice and is equal to 10 3 ( t + 3 ) 9 t 2 \dfrac{10}{3} (t + 3) \sqrt{9 - t^2} . The volume V V is half the volume of the cylinder, V = 45 π V = 45 \pi

Using the substitution t = 3 cos θ t = -3 \cos \theta , then d t = 3 sin θ d θ dt =3 \sin \theta d \theta , and the integral becomes:

G = ( 1 45 π ) θ = 0 π ( 3 cos θ , 0 , 5 2 ( 1 cos θ ) ) ( 90 ) ( 1 cos θ ) sin 2 θ d θ G = \displaystyle \left( \dfrac{1}{45 \pi} \right) \int_{\theta = 0}^{\pi} ( -3 \cos \theta, 0, \dfrac{5}{2} (1 - \cos \theta) ) \cdot (90) (1 - \cos \theta) \sin^2 \theta d \theta

Simplifying and expanding,

G = ( 2 π ) θ = 0 π ( 3 cos θ ( 1 cos θ ) sin 2 θ , 0 , 5 2 sin 2 θ ( 1 cos θ ) 2 ) d θ G = \displaystyle \left( \dfrac{2}{ \pi} \right) \int_{\theta = 0}^{\pi} ( -3 \cos \theta (1 - \cos \theta) \sin^2 \theta, 0, \dfrac{5}{2} \sin^2 \theta (1 - \cos \theta)^2 ) d \theta

Now,

θ = 0 π 3 cos θ ( 1 cos θ ) sin 2 θ d θ = θ = 0 π 3 cos 2 θ sin 2 θ d θ = ( 3 4 ) θ = 0 π sin 2 ( 2 θ ) d θ = ( 3 4 ) θ = 0 π 1 2 ( 1 cos ( 4 θ ) ) d θ = 3 π 8 \displaystyle \int_{\theta = 0}^{\pi} - 3 \cos \theta (1 - \cos \theta) \sin^2 \theta d \theta = \int_{\theta = 0}^{\pi} 3 \cos^2 \theta \sin^2 \theta d \theta = \left( \dfrac{3}{4} \right) \int_{\theta = 0}^{\pi} \sin^2 (2 \theta) d \theta = \left( \dfrac{3}{4} \right) \displaystyle \int_{\theta = 0}^{\pi} \dfrac{1}{2}(1- \cos (4 \theta)) d \theta = \dfrac{3 \pi }{8}

And,

θ = 0 π 5 2 sin 2 θ ( 1 cos θ ) 2 d θ = 5 2 θ = 0 π sin 2 θ 2 cos θ sin 2 θ + cos 2 θ sin 2 θ d θ \displaystyle \int_{\theta=0}^{\pi} \dfrac{5}{2} \sin^2 \theta (1 - \cos \theta)^2 d \theta = \dfrac{5}{2} \int_{\theta=0}^{\pi} \sin^2 \theta - 2 \cos \theta \sin^2 \theta + \cos^2 \theta \sin^2 \theta d \theta

= 5 2 ( π 2 0 + π 8 ) = ( 5 2 ) ( 5 π 8 ) = 25 π 16 = \dfrac{5}{2} ( \dfrac{\pi}{2} - 0 + \dfrac{\pi}{8} ) = \displaystyle \left( \dfrac{5}{2} \right) \left( \dfrac{5 \pi}{8} \right) = \dfrac{25 \pi}{16}

Therefore,

G = ( 2 π ) ( 3 π 8 , 0 , 25 π 16 ) = ( 3 4 , 0 , 25 8 ) G = \displaystyle \left( \dfrac{2}{\pi} \right) ( \dfrac{3 \pi}{8} , 0, \dfrac{25 \pi}{16} ) = ( \dfrac{3}{4} , 0, \dfrac{25}{8} )

So that x = 3 4 , y = 0 , z = 25 8 x = \dfrac{3}{4} , y = 0, z = \dfrac{25}{8} , making the answer 4 x + 8 z = 3 + 25 = 28 4 x + 8 z = 3 + 25 = \boxed{28}

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