Can we find the centre of mass of a solid hemisphere?
Question
The region in the first quadrant enclosed by the curve y = b ( a 2 − x 2 ) 1 / 2 a y = \dfrac{b(a^2 - x^2)^{1/2}}{a} y = a b ( a 2 − x 2 ) 1/2 y = 0 y = 0 y = 0 x = 0 x = 0 x = 0 x = k a x = ka x = k a ( 0 < k ≤ 1 ) (0 < k \le 1) ( 0 < k ≤ 1 ) x x x 1 3 π a b 2 k ( 3 − k 2 ) \tfrac{1}{3} \pi ab^2k(3-k^2) 3 1 π a b 2 k ( 3 − k 2 )
If this solid is of uniform density find the coordinates of its centre of mass.
By considering the case b = a b = a b = a k = 1 k = 1 k = 1 a a a 3 8 a \tfrac{3}{8}a 8 3 a
Solution
The volume of revolution of a curve around the
x x x
-axis is given by the expression
∫ π y 2 d x . \begin{equation*} \int \pi y^2 dx. \end{equation*} ∫ π y 2 d x . So the volume we need is
∫ 0 k a π b 2 a 2 − x 2 a 2 d x = π b 2 a 2 [ a 2 x − x 3 3 ] 0 k a = π b 2 a 2 ( k a 3 − k 3 a 3 3 ) = π b 2 k a ( 1 − k 2 3 ) = 1 3 π a b 2 k ( 3 − k 2 ) . \begin{align*} \int_0^{ka} \pi b^2 \frac{a^2 - x^2}{a^2} \:dx = \frac{\pi b^2}{a^2} \left[ a^2x - \frac{x^3}{3} \right]_0^{ka} &= \frac{\pi b^2}{a^2} \left( ka^3 - \frac{k^3a^3}{3} \right) \\ &= \pi b^2 ka \left( 1 - \frac{k^2}{3} \right) \\ &= \frac{1}{3} \pi ab^2k (3-k^2). \end{align*} ∫ 0 k a π b 2 a 2 a 2 − x 2 d x = a 2 π b 2 [ a 2 x − 3 x 3 ] 0 k a = a 2 π b 2 ( k a 3 − 3 k 3 a 3 ) = π b 2 k a ( 1 − 3 k 2 ) = 3 1 π a b 2 k ( 3 − k 2 ) . As this solid has rotational symmetry around the x x x x x x ( x ˉ , 0 ) (\bar{x},0) ( x ˉ , 0 )
The moment of our solid about an axis through O O O 1 1 1 1 3 π a b 2 k ( 3 − k 2 ) × x ˉ . \frac{1}{3}\pi ab^2k (3-k^2) \times \bar{x}. 3 1 π a b 2 k ( 3 − k 2 ) × x ˉ .
This must equal the sum of the moments of the infinitesimal cylinders we added in the first part of the question.
This is (in the limit)
∫ 0 k a π y 2 x d x = ∫ 0 k a π b 2 a 2 x − x 3 a 2 d x = π b 2 a 2 [ a 2 x 2 2 − x 4 4 ] 0 k a = π b 2 a 2 ( a 4 k 2 2 − a 4 k 4 4 ) = π a 2 b 2 k 2 4 ( 2 − k 2 ) . \begin{align*} \int_0^{ka} \pi y^2x \:dx &= \int_0^{ka} \pi b^2 \frac{a^2x - x^3}{a^2} \:dx\\ &= \frac{\pi b^2}{a^2} \left[ \dfrac{a^2x^2}{2} - \frac{x^4}{4} \right]_0^{ka}\\ &= \frac{\pi b^2}{a^2} \left( \dfrac{a^4k^2}{2} - \frac{a^4k^4}{4}\right)\\ &= \frac{\pi a^2b^2k^2}{4}(2 -k^2). \end{align*} ∫ 0 k a π y 2 x d x = ∫ 0 k a π b 2 a 2 a 2 x − x 3 d x = a 2 π b 2 [ 2 a 2 x 2 − 4 x 4 ] 0 k a = a 2 π b 2 ( 2 a 4 k 2 − 4 a 4 k 4 ) = 4 π a 2 b 2 k 2 ( 2 − k 2 ) . Thus
π a 2 b 2 k 2 4 ( 2 − k 2 ) = 1 3 π a b 2 k ( 3 − k 2 ) x ˉ ⟹ a k 4 ( 2 − k 2 ) = 1 3 ( 3 − k 2 ) x ˉ ⟹ 3 a k ( 2 − k 2 ) = 4 ( 3 − k 2 ) x ˉ ⟹ x ˉ = 3 a k ( 2 − k 2 ) 4 ( 3 − k 2 ) . \begin{align*} \frac{\pi a^2b^2k^2}{4}(2 -k^2)&=\frac{1}{3} \pi ab^2k (3-k^2)\bar{x}\\ \implies \frac{ak}{4}(2 -k^2)&=\frac{1}{3}(3-k^2)\bar{x}\\ \implies 3ak(2 -k^2)&=4(3-k^2)\bar{x}\\ \implies \bar{x} &= \dfrac{3ak(2 -k^2)}{4(3-k^2)}. \end{align*} 4 π a 2 b 2 k 2 ( 2 − k 2 ) ⟹ 4 ak ( 2 − k 2 ) ⟹ 3 ak ( 2 − k 2 ) ⟹ x ˉ = 3 1 π a b 2 k ( 3 − k 2 ) x ˉ = 3 1 ( 3 − k 2 ) x ˉ = 4 ( 3 − k 2 ) x ˉ = 4 ( 3 − k 2 ) 3 ak ( 2 − k 2 ) . By taking
b = a b = a b = a
and
k = 1 k = 1 k = 1
, we have that
y = a 2 − x 2 \begin{equation*} y = \sqrt{a^2 - x^2} \end{equation*} y = a 2 − x 2 and so
x 2 + y 2 = a 2 . \begin{equation*} x^2 + y^2 = a^2. \end{equation*} x 2 + y 2 = a 2 . This is the equation of a circle, centred at the origin and with radius a a a
Our area will rotate to a perfect hemisphere, since our integral’s limits are
0 0 0
and
a a a
, and so
x = k a x=ka x = k a
is tangent to the circle. Moreover,
x ˉ = 3 a k ( 2 − k 2 ) 4 ( 3 − k 2 ) = 3 a ( 2 − 1 2 ) 4 ( 3 − 1 2 ) = 3 a 8 , \begin{equation*} \bar{x} = \frac{3ak(2 - k^2)}{4(3-k^2)} = \frac{3a(2 - 1^2)}{4(3 - 1^2)} = \frac{3a}{8}, \end{equation*} x ˉ = 4 ( 3 − k 2 ) 3 ak ( 2 − k 2 ) = 4 ( 3 − 1 2 ) 3 a ( 2 − 1 2 ) = 8 3 a , and so the centre of mass of a uniform solid hemisphere of radius a a a 3 8 a \dfrac{3}{8}a 8 3 a
