Introduction
In this part we’ll cover how we change our variables into others, in the most general case.
General case of variable change
∬ D f ( x , y ) d A \iint_D f(x, y) \, dA ∬ D f ( x , y ) d A We want to rewrite this in terms of u u u v v v
x = g ( u , v ) y = h ( u , v ) \begin{align*}
x & = g(u, v) \newline
y & = h(u, v)
\end{align*} x y = g ( u , v ) = h ( u , v ) We can say that we have a transformation , T T T u v uv uv x y xy x y g g g h h h
We need to make some assumptions of T T T
T T T C 1 C^1 C 1 g g g h h h
T T T u u u v v v x x x y y y
With this, we can define the Jacobian matrix.
Definition 1 (Jacobian) The Jacobian matrix of a transformation, T T T
∂ ( x , y ) ∂ ( u , v ) = det [ ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ] \dfrac{\partial(x, y)}{\partial(u, v)} =
\det
\begin{bmatrix}
\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \newline
\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}
\end{bmatrix} ∂ ( u , v ) ∂ ( x , y ) = det ∂ u ∂ x ∂ u ∂ y ∂ v ∂ x ∂ v ∂ y Theorem 1 (Change of variables) Suppose T T T u v uv uv x y xy x y T T T
∬ D f ( x , y ) d A = ∬ S f ( x ( u , v ) , y ( u , v ) ) ∣ ∂ ( x , y ) ∂ ( u , v ) ∣ d u d v \iint_D f(x, y) \, dA = \iint_S f(x(u, v), y(u, v)) \left\vert \dfrac{\partial(x, y)}{\partial(u, v)} \right\vert \, du \, dv ∬ D f ( x , y ) d A = ∬ S f ( x ( u , v ) , y ( u , v )) ∂ ( u , v ) ∂ ( x , y ) d u d v Example 1 Find
∬ D e x + y x − y d A \iint_D e^{\frac{x + y}{x - y}} \, dA ∬ D e x − y x + y d A Where D D D ( 1 , 0 ) , ( 2 , 0 ) , ( 0 , − 2 ) (1, 0), (2, 0), (0, -2) ( 1 , 0 ) , ( 2 , 0 ) , ( 0 , − 2 ) ( 0 , − 1 ) (0, -1) ( 0 , − 1 )
Let u = x + y u = x + y u = x + y v = x − y v = x - y v = x − y
This means that:
x = u + v 2 y = u − v 2 \begin{align*}
x & = \frac{u + v}{2} \newline
y & = \frac{u - v}{2}
\end{align*} x y = 2 u + v = 2 u − v Let’s compute the Jacobian:
∂ ( x , y ) ∂ ( u , v ) = det [ ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ] \dfrac{\partial(x, y)}{\partial(u, v)} =
\det
\begin{bmatrix}
\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \newline
\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}
\end{bmatrix} ∂ ( u , v ) ∂ ( x , y ) = det ∂ u ∂ x ∂ u ∂ y ∂ v ∂ x ∂ v ∂ y ∂ ( x , y ) ∂ ( u , v ) = det [ 1 2 1 2 1 2 − 1 2 ] \dfrac{\partial(x, y)}{\partial(u, v)} =
\det
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2} \newline
\frac{1}{2} & -\frac{1}{2}
\end{bmatrix} ∂ ( u , v ) ∂ ( x , y ) = det [ 2 1 2 1 2 1 − 2 1 ] ∂ ( x , y ) ∂ ( u , v ) = − 1 2 \dfrac{\partial(x, y)}{\partial(u, v)} = -\frac{1}{2} ∂ ( u , v ) ∂ ( x , y ) = − 2 1 Since our transformation is linear, we only need to recompute our vertices. Using our formulas we find that
( 1 , 0 ) → ( 1 , 1 ) ( 2 , 0 ) → ( 2 , 2 ) ( 0 , − 2 ) → ( − 2 , 2 ) ( 0 , − 1 ) → ( − 1 , 1 ) (1, 0) \rarr (1, 1) \newline
(2, 0) \rarr (2, 2) \newline
(0, -2) \rarr (-2, 2) \newline
(0, -1) \rarr (-1, 1) ( 1 , 0 ) → ( 1 , 1 ) ( 2 , 0 ) → ( 2 , 2 ) ( 0 , − 2 ) → ( − 2 , 2 ) ( 0 , − 1 ) → ( − 1 , 1 ) If we plot our trapezoid, we find that it is a type II region. The left curve being u = − v u = -v u = − v u = v u = v u = v
Which means our integral becomes:
1 2 ∫ 1 2 ∫ − v v e u v d u d v \dfrac{1}{2} \int_1^2 \int_{-v}^{v} e^{\frac{u}{v}} \, du \, dv 2 1 ∫ 1 2 ∫ − v v e v u d u d v 1 2 ∫ 1 2 v e u v ∣ u = − v u = v d v \dfrac{1}{2} \int_1^2 ve^{\frac{u}{v}} \bigg\rvert_{u = -v}^{u = v} \, dv 2 1 ∫ 1 2 v e v u u = − v u = v d v 1 2 ( e − e − 1 ) ∫ 1 2 v d v \dfrac{1}{2} (e - e^{-1}) \int_1^2 v \, dv 2 1 ( e − e − 1 ) ∫ 1 2 v d v 1 2 ( e − e − 1 ) v 2 2 ∣ v = 1 v = 2 \dfrac{1}{2} (e - e^{-1}) \dfrac{v^2}{2} \bigg\rvert_{v = 1}^{v = 2} 2 1 ( e − e − 1 ) 2 v 2 v = 1 v = 2 3 ( e − e − 1 ) 4 \boxed{\dfrac{3(e - e^{-1})}{4}} 4 3 ( e − e − 1 ) In three variables
It is the same approach. Let x = x ( u , v , w ) , y = y ( u , v , w ) , z = z ( u , v , w ) x = x(u, v, w), y = y(u, v, w), z = z(u, v, w) x = x ( u , v , w ) , y = y ( u , v , w ) , z = z ( u , v , w )
The Jacobian is:
∂ ( x , y , z ) ∂ ( u , v , w ) = det [ ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w ] \dfrac{\partial(x, y, z)}{\partial(u, v, w)} =
\det
\begin{bmatrix}
\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \newline
\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \newline
\dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w}
\end{bmatrix} ∂ ( u , v , w ) ∂ ( x , y , z ) = det ∂ u ∂ x ∂ u ∂ y ∂ u ∂ z ∂ v ∂ x ∂ v ∂ y ∂ v ∂ z ∂ w ∂ x ∂ w ∂ y ∂ w ∂ z Just as we did before:
∭ E f ( x , y , z ) d V = ∭ S f ( x ( u , v , w ) , y ( u , v , w ) , z ( u , v , w ) ) ∣ ∂ ( x , y , z ) ∂ ( u , v , w ) ∣ d u d v d w \iiint_E f(x, y, z) \, dV = \iiint_S f(x(u, v, w), y(u, v, w), z(u, v, w)) \left\vert \dfrac{\partial(x, y, z)}{\partial(u, v, w)} \right\vert \, du \, dv \, dw ∭ E f ( x , y , z ) d V = ∭ S f ( x ( u , v , w ) , y ( u , v , w ) , z ( u , v , w )) ∂ ( u , v , w ) ∂ ( x , y , z ) d u d v d w Let’s now cover some special cases that we’ll often encounter.
Cylindrical coordinates
Is just like polar coordinates but for three variables:
( x , y , z ) → ( r , θ , z ) (x, y, z) \rarr (r, \theta, z) ( x , y , z ) → ( r , θ , z ) If we have symmetry around the z z z
We know that:
x = r cos ( θ ) y = r sin ( θ ) z = z \begin{align*}
x & = r \cos(\theta) \newline
y & = r \sin(\theta) \newline
z & = z
\end{align*} x y z = r cos ( θ ) = r sin ( θ ) = z Therefore, the Jacobian for this becomes:
∂ ( x , y , z ) ∂ ( r , θ , z ) = det [ cos ( θ ) − r sin ( θ ) 0 sin ( θ ) r cos ( θ ) 0 0 0 1 ] \dfrac{\partial(x, y, z)}{\partial(r, \theta, z)} =
\det
\begin{bmatrix}
\cos(\theta) & -r \sin(\theta) & 0 \newline
\sin(\theta) & r \cos(\theta) & 0 \newline
0 & 0 & 1
\end{bmatrix} ∂ ( r , θ , z ) ∂ ( x , y , z ) = det cos ( θ ) sin ( θ ) 0 − r sin ( θ ) r cos ( θ ) 0 0 0 1 ∂ ( x , y , z ) ∂ ( r , θ , z ) = 0 ⋅ det [ − r sin ( θ ) 0 r cos ( θ ) 0 ] − 0 ⋅ det [ cos ( θ ) 0 sin ( θ ) 0 ] + 1 ⋅ det [ cos ( θ ) − r sin ( θ ) sin ( θ ) r cos ( θ ) ] \dfrac{\partial(x, y, z)}{\partial(r, \theta, z)} = 0 \cdot \det
\begin{bmatrix}
-r \sin(\theta) & 0 \newline
r \cos(\theta) & 0
\end{bmatrix} - 0 \cdot \det
\begin{bmatrix}
\cos(\theta) & 0 \newline
\sin(\theta) & 0
\end{bmatrix} + 1 \cdot \det
\begin{bmatrix}
\cos(\theta) & -r \sin(\theta) \newline
\sin(\theta) & r \cos(\theta)
\end{bmatrix} ∂ ( r , θ , z ) ∂ ( x , y , z ) = 0 ⋅ det [ − r sin ( θ ) r cos ( θ ) 0 0 ] − 0 ⋅ det [ cos ( θ ) sin ( θ ) 0 0 ] + 1 ⋅ det [ cos ( θ ) sin ( θ ) − r sin ( θ ) r cos ( θ ) ] ∂ ( x , y , z ) ∂ ( r , θ , z ) = r cos 2 ( θ ) + r sin 2 ( θ ) = r ( cos 2 ( θ ) + sin 2 ( θ ) ) = r \dfrac{\partial(x, y, z)}{\partial(r, \theta, z)} = r \cos^2(\theta) + r \sin^2(\theta) = r(\cos^2(\theta) + \sin^2(\theta)) = \boxed{r} ∂ ( r , θ , z ) ∂ ( x , y , z ) = r cos 2 ( θ ) + r sin 2 ( θ ) = r ( cos 2 ( θ ) + sin 2 ( θ )) = r Change to spherical coordinates
∭ E f ( x , y , z ) d V = ∭ S f ( r cos ( θ ) , r sin ( θ ) , z ) ⋅ r d V \iiint_E f(x, y, z) \, dV = \iiint_S f(r \cos(\theta), r \sin(\theta), z) \cdot r \, dV ∭ E f ( x , y , z ) d V = ∭ S f ( r cos ( θ ) , r sin ( θ ) , z ) ⋅ r d V Example 2 Find
∭ E x 2 + y 2 d V \iiint_E \sqrt{x^2 + y^2} \, dV ∭ E x 2 + y 2 d V Where E E E x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1 z = 4 z = 4 z = 4 z = 1 − x 2 − y 2 z = 1 - x^2 - y^2 z = 1 − x 2 − y 2
Let us describe the solid E E E
E = { ( x , y , z ) ∣ 0 ≤ x 2 + y 2 ≤ 1 , 1 − x 2 − y 2 ≤ z ≤ 4 } E = \{(x, y, z) \mid 0 \leq x^2 + y^2 \leq 1, 1 - x^2 - y^2 \leq z \leq 4\} E = {( x , y , z ) ∣ 0 ≤ x 2 + y 2 ≤ 1 , 1 − x 2 − y 2 ≤ z ≤ 4 } In spherical coordinates:
S = { ( r , θ , z ) ∣ 0 ≤ r ≤ 1 , 1 − r 2 ≤ z ≤ 4 } S = \{(r, \theta, z) \mid 0 \leq r \leq 1, 1 - r^2 \leq z \leq 4\} S = {( r , θ , z ) ∣ 0 ≤ r ≤ 1 , 1 − r 2 ≤ z ≤ 4 } Let’s make z z z
∭ E x 2 + y 2 d V = ∭ S r 2 ⋅ r d V \iiint_E \sqrt{x^2 + y^2} \, dV = \iiint_S \sqrt{r^2} \cdot r \, dV ∭ E x 2 + y 2 d V = ∭ S r 2 ⋅ r d V ∭ S r 2 d V \iiint_S r^2 \, dV ∭ S r 2 d V ∫ 0 2 π ∫ 0 1 ∫ 1 − r 2 4 r 2 d z d r d θ \int_0^{2\pi} \int_0^1 \int_{1 - r^2}^{4} r^2 \, dz \, dr \, d\theta ∫ 0 2 π ∫ 0 1 ∫ 1 − r 2 4 r 2 d z d r d θ ∫ 0 2 π ∫ 0 1 r 2 z ∣ z = 1 − r 2 z = 4 d r d θ \int_0^{2\pi} \int_0^1 r^2 z \bigg\rvert_{z = 1 - r^2}^{z = 4} \, dr \, d\theta ∫ 0 2 π ∫ 0 1 r 2 z z = 1 − r 2 z = 4 d r d θ ∫ 0 2 π ∫ 0 1 ( 3 r 2 + r 4 ) d r d θ \int_0^{2\pi} \int_0^1 (3r^2 + r^4) \, dr \, d\theta ∫ 0 2 π ∫ 0 1 ( 3 r 2 + r 4 ) d r d θ ∫ 0 2 π ( r 3 + r 5 5 ) ∣ r = 0 r = 1 d θ \int_0^{2\pi} \left(r^3 + \dfrac{r^5}{5}\right) \bigg\rvert_{r = 0}^{r = 1} \, d\theta ∫ 0 2 π ( r 3 + 5 r 5 ) r = 0 r = 1 d θ ∫ 0 2 π 6 5 d θ \int_0^{2\pi} \dfrac{6}{5} \, d\theta ∫ 0 2 π 5 6 d θ 6 5 θ ∣ θ = 0 θ = 2 π \dfrac{6}{5}\theta \bigg\rvert_{\theta = 0}^{\theta = 2\pi} 5 6 θ θ = 0 θ = 2 π 12 π 5 \boxed{\dfrac{12\pi}{5}} 5 12 π Spherical coordinates
In this coordinate system we introduce a completely new idea.
Let:
ρ − distance from origin φ − angle between positive z -axis θ − angle in x y -plane \begin{align*}
\rho & - \text{distance from origin} \newline
\varphi & - \text{angle between positive } z\text{-axis} \newline
\theta & - \text{angle in } xy\text{-plane}
\end{align*} ρ φ θ − distance from origin − angle between positive z -axis − angle in x y -plane This means:
ρ ≥ 0 0 ≤ φ ≤ π 0 ≤ θ ≤ 2 π \rho \geq 0 \newline
0 \leq \varphi \leq \pi \newline
0 \leq \theta \leq 2\pi ρ ≥ 0 0 ≤ φ ≤ π 0 ≤ θ ≤ 2 π r = ρ sin ( φ ) x = r cos ( θ ) = ρ sin ( φ ) cos ( θ ) y = r sin ( θ ) = ρ sin ( φ ) sin ( θ ) z = ρ cos ( φ ) \begin{align*}
r & = \rho \sin(\varphi) \newline
x & = r \cos(\theta) = \rho \sin(\varphi) \cos(\theta) \newline
y & = r \sin(\theta) = \rho \sin(\varphi) \sin(\theta) \newline
z & = \rho \cos(\varphi)
\end{align*} r x y z = ρ sin ( φ ) = r cos ( θ ) = ρ sin ( φ ) cos ( θ ) = r sin ( θ ) = ρ sin ( φ ) sin ( θ ) = ρ cos ( φ ) The Jacobian:
∂ ( x , y , z ) ∂ ( ρ , φ , θ ) = det [ sin ( φ ) cos ( θ ) ρ cos ( φ ) cos ( θ ) − ρ sin ( φ ) sin ( θ ) sin ( φ ) sin ( θ ) ρ cos ( φ ) sin ( θ ) ρ sin ( φ ) cos ( θ ) cos ( φ ) − ρ sin ( φ ) 0 ] \dfrac{\partial(x, y, z)}{\partial(\rho, \varphi, \theta)} =
\det
\begin{bmatrix}
\sin(\varphi) \cos(\theta) & \rho \cos(\varphi) \cos(\theta) & -\rho \sin(\varphi) \sin(\theta) \newline
\sin(\varphi) \sin(\theta) & \rho \cos(\varphi) \sin(\theta) & \rho \sin(\varphi) \cos(\theta) \newline
\cos(\varphi) & -\rho \sin(\varphi) & 0
\end{bmatrix} ∂ ( ρ , φ , θ ) ∂ ( x , y , z ) = det sin ( φ ) cos ( θ ) sin ( φ ) sin ( θ ) cos ( φ ) ρ cos ( φ ) cos ( θ ) ρ cos ( φ ) sin ( θ ) − ρ sin ( φ ) − ρ sin ( φ ) sin ( θ ) ρ sin ( φ ) cos ( θ ) 0 ∂ ( x , y , z ) ∂ ( ρ , φ , θ ) = … = ρ 2 sin ( φ ) \dfrac{\partial(x, y, z)}{\partial(\rho, \varphi, \theta)} = \ldots = \rho^2 \sin(\varphi) ∂ ( ρ , φ , θ ) ∂ ( x , y , z ) = … = ρ 2 sin ( φ ) Changing to spherical coordinates
∭ E f ( x , y , z ) d V = ∭ S f ( ρ sin ( φ ) cos ( θ ) , ρ sin ( φ ) sin ( θ ) , ρ cos ( φ ) ) ⋅ ρ 2 sin ( φ ) d V \iiint_E f(x, y, z) \, dV = \iiint_S f(\rho \sin(\varphi) \cos(\theta), \rho \sin(\varphi) \sin(\theta), \rho \cos(\varphi)) \cdot \rho^2 \sin(\varphi) \, dV ∭ E f ( x , y , z ) d V = ∭ S f ( ρ sin ( φ ) cos ( θ ) , ρ sin ( φ ) sin ( θ ) , ρ cos ( φ )) ⋅ ρ 2 sin ( φ ) d V It makes sense to change to spherical coordinates when our solid is bounded by spheres and/or cones.
Example 3 Find
∭ B e ( x 2 + y 2 + z 2 ) 3 2 d V \iiint_B e^{(x^2 + y^2 + z^2)^{\frac{3}{2}}} \, dV ∭ B e ( x 2 + y 2 + z 2 ) 2 3 d V Where B B B
B = { ( x , y , z ) ∣ x 2 + y 2 + z 2 ≤ 1 } B = \{(x, y, z) \mid x^2 + y^2 + z^2 \leq 1 \} B = {( x , y , z ) ∣ x 2 + y 2 + z 2 ≤ 1 } In spherical coordinates:
S = { ( ρ , φ , θ ) ∣ 0 ≤ ρ ≤ 1 , 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2 π } S = \{(\rho, \varphi, \theta) \mid 0 \leq \rho \leq 1, 0 \leq \varphi \leq \pi, 0 \leq \theta \leq 2\pi \} S = {( ρ , φ , θ ) ∣ 0 ≤ ρ ≤ 1 , 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2 π } Which is easy to integrate over.
∭ B e ( x 2 + y 2 + z 2 ) 3 2 d V = ∭ S e ( ρ 2 ) 3 2 ⋅ ρ 2 sin ( φ ) d V \iiint_B e^{(x^2 + y^2 + z^2)^{\frac{3}{2}}} \, dV = \iiint_S e^{(\rho^2)^{\frac{3}{2}}} \cdot \rho^2 \sin(\varphi) \, dV ∭ B e ( x 2 + y 2 + z 2 ) 2 3 d V = ∭ S e ( ρ 2 ) 2 3 ⋅ ρ 2 sin ( φ ) d V ∭ S e ρ 3 ⋅ ρ 2 sin ( φ ) d V \iiint_S e^{\rho^3} \cdot \rho^2 \sin(\varphi) \, dV ∭ S e ρ 3 ⋅ ρ 2 sin ( φ ) d V ∫ 0 2 π ∫ 0 1 ∫ 0 π e ρ 3 ⋅ ρ 2 sin ( φ ) d φ d ρ d θ \int_0^{2\pi} \int_0^1 \int_0^{\pi} e^{\rho^3} \cdot \rho^2 \sin(\varphi) \, d\varphi \, d\rho \, d\theta ∫ 0 2 π ∫ 0 1 ∫ 0 π e ρ 3 ⋅ ρ 2 sin ( φ ) d φ d ρ d θ ∫ 0 2 π ∫ 0 1 e ρ 3 ⋅ ρ 2 [ − cos ( φ ) ∣ φ = 0 φ = π ] d ρ d θ \int_0^{2\pi} \int_0^1 e^{\rho^3} \cdot \rho^2 \left[-\cos(\varphi) \bigg\rvert_{\varphi = 0}^{\varphi = \pi} \right] \, d\rho \, d\theta ∫ 0 2 π ∫ 0 1 e ρ 3 ⋅ ρ 2 [ − cos ( φ ) φ = 0 φ = π ] d ρ d θ ∫ 0 2 π ∫ 0 1 2 e ρ 3 ⋅ ρ 2 d ρ d θ \int_0^{2\pi} \int_0^1 2e^{\rho^3} \cdot \rho^2 \, d\rho \, d\theta ∫ 0 2 π ∫ 0 1 2 e ρ 3 ⋅ ρ 2 d ρ d θ Let
u = ρ 3 u = \rho^3 u = ρ 3 d u = 3 ρ 2 d ρ du = 3 \rho^2 \, d\rho d u = 3 ρ 2 d ρ d u 3 = ρ 2 d ρ \frac{du}{3} = \rho^2 \, d\rho 3 d u = ρ 2 d ρ ∫ 0 2 π ∫ 0 1 2 3 e u d u d θ \int_0^{2\pi} \int_0^1 \frac{2}{3} e^u \, du \, d\theta ∫ 0 2 π ∫ 0 1 3 2 e u d u d θ ∫ 0 2 π 2 3 e u ∣ u = 0 u = 1 d θ \int_0^{2\pi} \frac{2}{3} e^u \bigg\rvert_{u = 0}^{u = 1} \, d\theta ∫ 0 2 π 3 2 e u u = 0 u = 1 d θ ∫ 0 2 π 2 ( e − 1 ) 3 d θ \int_0^{2\pi} \frac{2(e - 1)}{3} \, d\theta ∫ 0 2 π 3 2 ( e − 1 ) d θ 2 ( e − 1 ) 3 θ ∣ θ = 0 θ = 2 π \frac{2(e - 1)}{3} \theta \bigg\rvert_{\theta = 0}^{\theta = 2\pi} 3 2 ( e − 1 ) θ θ = 0 θ = 2 π 4 π ( e − 1 ) 3 \boxed{\frac{4\pi(e - 1)}{3}} 3 4 π ( e − 1 ) Some applications of multiple integrals
Let’s see what we can achieve with multiple integrals.
Computing mass of solid
Let E E E ρ ( x , y , z ) \rho(x, y, z) ρ ( x , y , z )
We say that the mass of the solid is:
∭ E ρ ( x , y , z ) d V \iiint_E \rho(x, y, z) \, dV ∭ E ρ ( x , y , z ) d V We perform the same proof as for integrals for solids, but with the density function instead. Taking the sum and limit of small points and their densities.
Center of Mass
One very important application is finding the center of mass of a solid.
Given a physical system with m i m_i m i ( x 0 , y 0 , z 0 ) (x_0, y_0, z_0) ( x 0 , y 0 , z 0 )
Therefore:
x 0 = ∑ m i x i ∑ m i x_0 = \dfrac{\sum m_i x_i}{\sum m_i} x 0 = ∑ m i ∑ m i x i y 0 = ∑ m i y i ∑ m i y_0 = \dfrac{\sum m_i y_i}{\sum m_i} y 0 = ∑ m i ∑ m i y i z 0 = ∑ m i z i ∑ m i z_0 = \dfrac{\sum m_i z_i}{\sum m_i} z 0 = ∑ m i ∑ m i z i For an arbitrary solid, it’s almost the same:
x 0 = ∭ E ρ ( x , y , z ) ⋅ x d V ∭ E ρ ( x , y , z ) d V x_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot x \, dV}{\iiint_E \rho(x, y, z) \, dV} x 0 = ∭ E ρ ( x , y , z ) d V ∭ E ρ ( x , y , z ) ⋅ x d V y 0 = ∭ E ρ ( x , y , z ) ⋅ y d V ∭ E ρ ( x , y , z ) d V y_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot y \, dV}{\iiint_E \rho(x, y, z) \, dV} y 0 = ∭ E ρ ( x , y , z ) d V ∭ E ρ ( x , y , z ) ⋅ y d V z 0 = ∭ E ρ ( x , y , z ) ⋅ z d V ∭ E ρ ( x , y , z ) d V z_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot z \, dV}{\iiint_E \rho(x, y, z) \, dV} z 0 = ∭ E ρ ( x , y , z ) d V ∭ E ρ ( x , y , z ) ⋅ z d V Again, the proof is the exact same as the previous one.
Example 4 A rectangular solid of width 2, length 10 and height 2 has the density function, ρ ( x , y , z ) = y 3 \rho(x, y, z) = y^3 ρ ( x , y , z ) = y 3 x , y , z x, y, z x , y , z
So, we need to compute the center of mass of the solid. If we think logically about this, the only coordinate that matters is the y y y
If y 0 ≤ 2 3 ⋅ 10 y_0 \leq \dfrac{2}{3} \cdot 10 y 0 ≤ 3 2 ⋅ 10 y 0 y_0 y 0
y 0 = ∭ E ρ ( x , y , z ) ⋅ y d V m y_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot y \, dV}{m} y 0 = m ∭ E ρ ( x , y , z ) ⋅ y d V y 0 = ∭ E y 3 ⋅ y d V m y_0 = \dfrac{\iiint_E y^3 \cdot y \, dV}{m} y 0 = m ∭ E y 3 ⋅ y d V y 0 = ∭ E y 4 d V m y_0 = \dfrac{\iiint_E y^4 \, dV}{m} y 0 = m ∭ E y 4 d V Since we have a rectangular box:
y 0 = ∫ 0 2 ∫ 0 2 ∫ 0 10 y 4 d y d x d z m y_0 = \dfrac{\int_0^2 \int_0^2 \int_0^{10} y^4 \, dy \, dx \, dz}{m} y 0 = m ∫ 0 2 ∫ 0 2 ∫ 0 10 y 4 d y d x d z y 0 = ∫ 0 2 ∫ 0 2 10 5 5 d x d z m y_0 = \dfrac{\int_0^2 \int_0^2 \dfrac{10^5}{5} \, dx \, dz}{m} y 0 = m ∫ 0 2 ∫ 0 2 5 1 0 5 d x d z y 0 = ∫ 0 2 10 5 5 ⋅ 2 d z m y_0 = \dfrac{\int_0^2 \dfrac{10^5}{5} \cdot 2 \, dz}{m} y 0 = m ∫ 0 2 5 1 0 5 ⋅ 2 d z y 0 = 10 5 5 ⋅ 2 ⋅ 2 m y_0 = \dfrac{\dfrac{10^5}{5} \cdot 2 \cdot 2}{m} y 0 = m 5 1 0 5 ⋅ 2 ⋅ 2 Now let’s compute m m m
m = ∭ E ρ ( x , y , z ) d V m = \iiint_E \rho(x, y, z) \, dV m = ∭ E ρ ( x , y , z ) d V m = ∭ E y 3 d V m = \iiint_E y^3 \, dV m = ∭ E y 3 d V m = ∫ 0 2 ∫ 0 2 ∫ 0 10 y 3 d y d x d z m = \int_0^2 \int_0^2 \int_0^{10} y^3 \, dy \, dx \, dz m = ∫ 0 2 ∫ 0 2 ∫ 0 10 y 3 d y d x d z m = ∫ 0 2 ∫ 0 2 10 4 4 d x d z m = \int_0^2 \int_0^2 \dfrac{10^4}{4} \, dx \, dz m = ∫ 0 2 ∫ 0 2 4 1 0 4 d x d z m = ∫ 0 2 10 4 4 ⋅ 2 d z m = \int_0^2 \dfrac{10^4}{4} \cdot 2 \, dz m = ∫ 0 2 4 1 0 4 ⋅ 2 d z m = 10 4 4 ⋅ 2 ⋅ 2 m = \dfrac{10^4}{4} \cdot 2 \cdot 2 m = 4 1 0 4 ⋅ 2 ⋅ 2 y 0 = 10 5 5 ⋅ 2 ⋅ 2 10 4 4 ⋅ 2 ⋅ 2 = … = 8 y_0 = \dfrac{\dfrac{10^5}{5} \cdot 2 \cdot 2}{\dfrac{10^4}{4} \cdot 2 \cdot 2} = \ldots = \boxed{8} y 0 = 4 1 0 4 ⋅ 2 ⋅ 2 5 1 0 5 ⋅ 2 ⋅ 2 = … = 8