Examples from Mass, centroids and probability All examples from Mass, centroids and probability.
[ EXAMPLE ] Let
Ω = { ( x , y ) : ∣ x ∣ ≤ y ≤ 1 } , ρ ( x , y ) = x 2 + y 2 . \Omega=\{(x,y):|x|\le y\le1\},
\qquad
\rho(x,y)=x^2+y^2. Ω = {( x , y ) : ∣ x ∣ ≤ y ≤ 1 } , ρ ( x , y ) = x 2 + y 2 . Horizontal slices give
0 ≤ y ≤ 1 , − y ≤ x ≤ y . 0\le y\le1,\qquad-y\le x\le y. 0 ≤ y ≤ 1 , − y ≤ x ≤ y . The mass is
M = ∫ 0 1 ∫ − y y ( x 2 + y 2 ) d x d y = ∫ 0 1 [ x 3 3 + x y 2 ] − y y d y = ∫ 0 1 8 y 3 3 d y = 2 3 . \begin{aligned}
M
&=\int_0^1\int_{-y}^{y}(x^2+y^2)\,dx\,dy\\
&=\int_0^1\left[\frac{x^3}{3}+xy^2\right]_{-y}^{y}dy\\
&=\int_0^1\frac{8y^3}{3}\,dy
=\frac23.
\end{aligned} M = ∫ 0 1 ∫ − y y ( x 2 + y 2 ) d x d y = ∫ 0 1 [ 3 x 3 + x y 2 ] − y y d y = ∫ 0 1 3 8 y 3 d y = 3 2 . The region and density are symmetric under x ↦ − x x\mapsto-x x ↦ − x
x ˉ = 0. \bar x=0. x ˉ = 0. The vertical moment is
∬ Ω y ρ d A = ∫ 0 1 ∫ − y y y ( x 2 + y 2 ) d x d y = ∫ 0 1 8 y 4 3 d y = 8 15 . \begin{aligned}
\iint_\Omega y\rho\,dA
&=\int_0^1\int_{-y}^{y}y(x^2+y^2)\,dx\,dy\\
&=\int_0^1\frac{8y^4}{3}\,dy
=\frac{8}{15}.
\end{aligned} ∬ Ω y ρ d A = ∫ 0 1 ∫ − y y y ( x 2 + y 2 ) d x d y = ∫ 0 1 3 8 y 4 d y = 15 8 . Thus
y ˉ = 8 / 15 2 / 3 = 4 5 , \bar y=\frac{8/15}{2/3}=\frac45, y ˉ = 2/3 8/15 = 5 4 , and the centre of mass is
( 0 , 4 5 ) . {\left(0,\frac45\right)}. ( 0 , 5 4 ) . For uniform density, this triangle's centroid has y y y 2 / 3 2/3 2/3 y y y
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For uniform density ρ = 1 \rho=1 ρ = 1
M = π 2 . M=\frac{\pi}{2}. M = 2 π . The vertical moment is
∬ Ω y d A = ∫ 0 π ∫ 0 1 ( r sin θ ) r d r d θ = ( ∫ 0 π sin θ d θ ) ( ∫ 0 1 r 2 d r ) = 2 3 . \begin{aligned}
\iint_\Omega y\,dA
&=\int_0^\pi\int_0^1
(r\sin\theta)r\,dr\,d\theta\\
&=\left(\int_0^\pi\sin\theta\,d\theta\right)
\left(\int_0^1r^2\,dr\right)
=\frac23.
\end{aligned} ∬ Ω y d A = ∫ 0 π ∫ 0 1 ( r sin θ ) r d r d θ = ( ∫ 0 π sin θ d θ ) ( ∫ 0 1 r 2 d r ) = 3 2 . Therefore
y ˉ = 4 3 π . \bar y=\frac{4}{3\pi}. y ˉ = 3 π 4 . For ρ = y \rho=y ρ = y
M = ∬ Ω y d A = ∫ 0 π ∫ 0 1 r 2 sin θ d r d θ = 2 3 . \begin{aligned}
M
&=\iint_\Omega y\,dA\\
&=\int_0^\pi\int_0^1r^2\sin\theta\,dr\,d\theta
=\frac23.
\end{aligned} M = ∬ Ω y d A = ∫ 0 π ∫ 0 1 r 2 sin θ d r d θ = 3 2 . The vertical moment becomes
∬ Ω y ρ d A = ∬ Ω y 2 d A = ∫ 0 π ∫ 0 1 r 3 sin 2 θ d r d θ = π 8 . \begin{aligned}
\iint_\Omega y\rho\,dA
&=\iint_\Omega y^2\,dA\\
&=\int_0^\pi\int_0^1
r^3\sin^2\theta\,dr\,d\theta
=\frac{\pi}{8}.
\end{aligned} ∬ Ω y ρ d A = ∬ Ω y 2 d A = ∫ 0 π ∫ 0 1 r 3 sin 2 θ d r d θ = 8 π . Hence
y ˉ = π / 8 2 / 3 = 3 π 16 . \bar y=\frac{\pi/8}{2/3}=\frac{3\pi}{16}. y ˉ = 2/3 π /8 = 16 3 π . For radial density
ρ = x 2 + y 2 = r , \rho=\sqrt{x^2+y^2}=r, ρ = x 2 + y 2 = r , the mass is
M = ∫ 0 π ∫ 0 1 r ⋅ r d r d θ = π 3 , \begin{aligned}
M
&=\int_0^\pi\int_0^1r\cdot r\,dr\,d\theta
=\frac{\pi}{3},
\end{aligned} M = ∫ 0 π ∫ 0 1 r ⋅ r d r d θ = 3 π , and the vertical moment is
∬ Ω y ρ d A = ∫ 0 π ∫ 0 1 ( r sin θ ) r r d r d θ = 1 2 . \begin{aligned}
\iint_\Omega y\rho\,dA
&=\int_0^\pi\int_0^1
(r\sin\theta)\,r\,r\,dr\,d\theta
=\frac12.
\end{aligned} ∬ Ω y ρ d A = ∫ 0 π ∫ 0 1 ( r sin θ ) r r d r d θ = 2 1 . Thus
y ˉ = 1 / 2 π / 3 = 3 2 π . \bar y=\frac{1/2}{\pi/3}=\frac{3}{2\pi}. y ˉ = π /3 1/2 = 2 π 3 . The three heights are approximately 0.424 0.424 0.424 0.589 0.589 0.589 0.477 0.477 0.477 y y y
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