Multiple Integrations

My personal notes for more complex forms of integration.

Some equations might not show on mobile.

15.1 Integration in Two Variables

Double Integral

Integration of two variables

$$\iint_D f(x,y) dA$$

• Volume of region between graph when $f(x,y) \ge 0$ on domain $D$

Similarities Between Double and Single Integrals:

• Double Integrals are defined as limits of Riemann sums.
• Double integrals are evaluated using the Fundamental Theorem of Calculus, but applied twice

Integrals are broken into 3 parts

1. Subdivision
   Subdivide $[a,b]$ and $[c,d]$ by choosing partitions:
2. $$a=x_0 < x_1 < … <x_N=b, \quad c = y_0<y_1<…<y_M=d$$
3. Summation
   Create an $N \times M$ grid of sub-rectangles $R_{ij}$
4. Passage to the limit
   Choose a sample point $P_{ij}$ in each $R_{ij}$

Limit of Riemann Sums

The Riemann sum $S_{N,M}$ approaches a limit $L$ as $||P|| \rightarrow 0$ if, for all $\epsilon$ there exists $\delta > 0$ such that

$$|L-S_{N,M}| < \epsilon$$

for all partitions satisfying $||P||<\delta$ and all choices of sample points. We write

$$\lim_{||P|| \rightarrow 0} S_{N,M}= \lim_{||P|| \rightarrow 0}\sum^N_{i=1}\sum^M_{j=1}f(P_{ij})\Delta A_{ij}=L$$

where

$$\Delta x = \frac{b-a}{N}, \quad \Delta y = \frac{d-c}{M}$$

Double Integral over a Rectangle

The double integral of $f(x,y)$ over a rectangle $R$ is defined as the limit. This technique is used to find the volume of an area over a rectangle

$$\iint_R f(x,y)dA=\lim_{||P|| \rightarrow 0} \sum^N_{i=1} \sum^M_{j=1} f(P_{ij})\Delta A_{ij}$$

Example :: Estimating Double Integral:

Compute the Riemann sum $S_{4,3}$ to estimate the double integral of $f(x,y)=6xy$ over $R=[1,5]×[1,4]$. Use the regular partition and upper‑right vertices of the sub‑rectangles as sample points.

1. Calculate length of sides of subrectangle
   $$\Delta x = \frac{5-1}{4}=1, \quad \Delta y = \frac{4-1}{3}=1, \quad \Delta A= \Delta x \Delta y = 1$$
2. Plug into the corresponding Riemann sum
   $$S_{4,3} = \sum^N_{i=1} \sum^M_{j=1} f(P_{ij})\Delta A_{ij}= \sum^4_{i=1}\sum^3_{j=1} (6ij)\cdot (1)$$
3. Sum for Regular Partition
   

Theorem 1: Continuous Functions Are Integrable

If a function $f$ of two variables is continuous on a rectangle $R$, then $f(x,y)$ is integrable over $R$

Continuous

A function is continuous when its graph is a single unbroken curve

Theorem 2: Linearity of the Double Integral

Assume that $f(x,y)$ and $g(x,y)$ are integrable over a rectangle $R$. Then,
$$\text{i. } \iint_R (f(x,y) + g(x,y))dA = \iint_R f(x,y)dA + \iint_R g(x,y) dA$$
$$\text{ii. For any constant C…} \iint_R C \cdot f(x,y)dA= C \iint_R f(x,y)dA$$

Example :: Calculating a iterated integral:

Evaluate the iterated integral…

$$\int^5_1 \int^{\pi / 3} _0 x^2 \sin(y)dy dx$$

1. Take the inner integral (x)
2. $$\int^{\pi / 3}_{x=0} x^2 \cdot \sin(y)dy =\quad x^2 \cdot [-\cos(y)]|^{\pi / 3}_0=x^2(-\cos(\frac{\pi}{3})-\cos(0))=x^2 \cdot \frac{1}{2}$$
3. Take the outer integral
4. $$\frac{1}{2}\int^5_1 x^2dx = \frac{1}{2}[\frac{x^3}{3}]|^5_1=\frac{1}{2}(\frac{125}{3}-\frac{1}{3})=\frac{62}{3}$$

Example :: Double Integral:

Evaluate the integral
$$\iint _R \frac{23x}{y}dA, \quad R=[-2,4] \times [1,3]$$

1. Replace with more readable terms
2. $$\iint_R \frac{23x}{y}dA = \int{x=(-2)}^4 \int^3_1 \frac{23x}{y}dydx$$
3. Compute the inner integral
   $$\int^3_1 \frac{23x}{y}dy=23x\ln|y||^3_1=23x(\ln|3| - \ln |1|)$$
4. Compute the outer integral
   $$(\ln|3| - \ln |1|)\int_{x=(-2)}^4 23x=(\ln|3| - \ln |1|)\cdot (\frac{23x^2}{2})|^4_{-2}=…$$
5. $$(\ln|3| - \ln |1|) \cdot (\frac{23(4)^2}{2}-\frac{23(-2)^2}{2})$$

Evaluate the integral
$$\iint_R y^2 \tan (x) dA,R=[0,\frac{\pi}{3}] \times [0,5]$$

1. Rewrite integral above
2. $$\iint_R y^2 \tan (x) dA=\int^{\pi/3}_0 \int^5_0 y^2\tan(x)dydx$$
3. Evaluate the inner integral
4. $$\tan(x)\int^5_0 y^2dy=\tan(x)(\frac{y^3}{3})|^5_0=\tan(x)(\frac{5^3}{3}-0)=\frac{125}{3}\tan(x)$$
5. Evaluate the outer integral
6. $$\frac{125}{3} \int^{\pi/3}_0 \tan(x)=-(\frac{125}{3})(\ln|\cos(x)|)|^{\pi/3}_0=-(\frac{125}{3})(\ln|\cos(\pi/3)|-\ln|\cos(0)|)$$
7. Compute
8. $$-(\frac{125}{3})(\ln|\cos(\pi/3)|-\ln|\cos(0)|)=\frac{125}{3}\ln \left(2\right)$$

15.2 Double Integrals Over More General Regions

Piecewise Smooth

A multidimensional curve that acts as a boundary typically for integrals

Defining Double Integral of a Piecewise Smooth

The function $\tilde{f}$ is set equal to the function mapping the domain

$$\iint_D f(x,y)dA = \iint_R \tilde{f}(x,y)dA$$

$$\tilde{f}(x,y)= {f(x,y) \text{ if } (x,y) \in D \quad \& \quad 0 \text{ if } (x,y) \not\in D}$$

• function $f$ is integral over $D$ if the integral of $\tilde{f}$ over $R$ exists (as all shown in the image above)

Theorem 1

If $f(x,y)$ is continuous on a closed domain $D$ whose boundary is a simple closed piecewise smooth curve, then $\iint_D f(x,y)dA$ exists

Computing Regions Between Two Graphs

When $D$ is a region between two graphs in the xy-plane, we can evaluate double integrals over $D$ as interated integrals. Since $D$ is vertically simple if it is the region between the graph of two continuous functions $y=g_1(x)$ and $y=g_2(x)$ over a fixed interval.

$$D=\{(x,y):a \le x \le b, \quad g_1(x) \le y \le g_2 (x)\}$$

If $D$ is horizontally simple

$$D=\{(x,y):c \le y \le d, \quad h_1(y) \le x \le h_2 (y)\}$$

Theorem 2

If $D$ is vertically simple with description

$$D=\{(x,y):a \le x \le b, \quad g_1(x) \le y \le g_2 (x)\}$$

then,

$$\iint_D f(x,y)dA=\int^b_a \int^{g_2(x)}_{g_1(x)} f(x,y)dydx$$

If $D$ is horizontally simple with description

$$D=\{(x,y):c \le y \le d, \quad h_1(y) \le x \le h_2 (y)\}$$

then,

$$\iint_D f(x,y) dA=\int^d_c \int^{h_2(x)}_{h_1(x)} f(x,y)dxdy$$

Note:
Since $\tilde{f}(x,y)$ is zero outside $D$, so for fixed x, $\tilde{f}(x,y)$ is zero unless y satisfies $g_1(x) \le y \le g_2(x)$ giving us

$$\int^d_c \tilde{f}(x,y)dy = \int^{g_2(x)}_{g_1(x)} f(x,y)dydx$$

Integration over a horizontally or vertically simple region is similar to integration over a rectangle with one difference: The limits of the inner integral may be functions instead of constants.

Example :: Compute double integral over domain D

Compute the double integral of $f(x,y)=x^2y$ over the given shaded domain in the figure. Assume that $a=1$.

1. Describe $D$ as a vertically simple region.
   
   $$0 \le x \le (2a=2), \quad f(x)=? \le y \le a$$
2. We need to find the function $f(x)$ along the piecewise smooth. Knowing that we go right (2) a’s for every down (1) a we can create a function mapping to
   $$f(x)=\frac{y}{x}=a - \frac{1}{2}x=1- \frac{1}{2}x$$
   Now we can plug into our equation with value $a=1$ …
   $$0 \le x \le 2, \quad 1- \frac{1}{2}x \le y \le 1$$

This segment is horizontally simple since the x value doesn’t change in this piecewise smooth but the y-compontent does

3. Set up the iterated integral
4. $$\iint_D x^2ydA = \int^2_0 \int^1_{y=(1 - \frac{1}{2}x)}x^2y \space dydx$$
5. Solve for y on the inner integral
6. $$\int^1_{y=(1- \frac{1}{2}x)}x^2y \space dy=\int^2_0= \frac{1}{2}x^2y^2 \space \vert^1_{y=(1- \frac{1}{2}x)}= \int^2_0 x^2\left(\frac{1}{2}-\frac{\left(2-x\right)^2}{8}\right)dx$$
7. Solve for x on the outer integral with the new input value
8. $$\int^2_0 x^2\left(\frac{1}{2}-\frac{\left(2-x\right)^2}{8}\right)dx= \int^2_0 \frac{x^2}{2} - \frac{x^2(2-x)^2}{8} = 2 - \frac{4}{5}=1.2$$
   ANS: $1.2$

Compute the double integral of $\iint_D x^3y \space dA$ over the domain $D$ over $0 \le x \le 3, \quad x \le y \le 5x + 3$
Rewrite:
$$\int^3_0 \int^{5x+3}_x x^3 y \space dydx$$

1. Calculate the inner integral
   $$\int^{5x+3}_x x^3 y \space dy=\frac{1}{2}x^3y^2 \vert_x^{5x+3}=\frac{x^3(9+30x+24x^2)}{2}$$
2. Plug in the new value of the inner integral into the outer integral
   $$\int^3_0\frac{x^3\left(9+30x+24x^2\right)}{2}dx=\frac{18225}{8}$$
3. ANS: \dfrac{18225}{8} /)

Compute the double integral \(\iint_D (70xy-x^2) \space dA over the region bounded below by $y=x^2$, above by $y=\sqrt{x}$.

1. Find the bounds of the outer integral
   $$x^2 = \sqrt{x}, \quad x = 0,1$$
2. Rewrite the integral
3. $$\int^1_0 \int_{x^2}^{\sqrt{x}}(70xy-x^2) \space dydx$$
4. Compute the inner integral
5. $$\int_{x^2}^{\sqrt{x}}(70xy-x^2) \space dy=35xy^2-x^2 \vert_{x^2}^{\sqrt{x}}=-x^2\sqrt{x}+x^4-35x^5+35x^2$$
6. Compute the outer integral
7. $$\int^1_0(-x^2\sqrt{x}+x^4-35x^5+35x^2) \space dx=\frac{1207}{210}$$
   ANS: $\frac{1207}{210}$

Theorem 3

Let $f(x,y)$ and $g(x,y)$ be integrable functions on $D$

1. If $f(x,y) \le g(x,y)$ for all $(x,y) \in D$ then,
   • $$\iint_D f(x,y)dA \le \iint_D g(x,y)dA$$
2. If $m \le f(x,y) \le M$ for all $(x,y) \in D$ then
   • $$m \cdot area(D) \le \iint_D f(x,y)dA \le M \cdot area(D)$$

Average Value

The average (mean) value of a function on domain $D$ is denoted as $\overline{f}$

$$\overline{f}=\frac{1}{area(D)} \iint_D f(x,y)dA=\frac{\iint_D f(x,y)dA}{\iint_D 1dA}$$

The value of $\overline{f}$ satisfies this relationship

$$\iint_D f(x,y)dA=\overline{f} \cdot area(D)$$

Theorem 4: Mean Value Theorem for Double Integrals

If $f(x,y)$ is continuous and $D$ is closed, bounded, and connected, then there exists a point $P \in D$ such that

$$\iint_D f(x,y)dA=f(P)area(D)$$

Equivalently, $f(P)=\overline{f}$ where $\overline{f}$ is the average value of $f$ on $D$.

Note:
The mean value theorem for double integrals only applies to continuous functions, therefore meaning they lie on the same domain

Example :: Find volume from 3 points in 3D space:

Calculate the double integral of $f(x,y)=5-8x$ over the triangle with vertices $O=(0,0),A=(2,7),B=(6,7)$
$$\iint_D(5-8x)dA$$
$$\int^7_0 \int_{\frac{2}{7}y}^{\frac{6}{7}y}(5-8x)dxdy=-\frac{686}{3}$$

$$\int^6_0 \int^{7x/2}_{6x/2}(5-8x)dxdy$$

Find the volume of the region bounded by

$$z=72-y,\quad z=y,\quad y=x^2, \quad y=36-x^2$$
$$z=72-(z),\quad z=36, \quad x^2=36-x^2, \quad x=\sqrt{18}, \quad y=18$$

Example :: Estimating area of the subdomain:

Using the table below find the average area of the list of sub-domains

Estimate $\iint_D f(x,y) dA$

$$\iint_D f(x,y)dA=f(P)area(D)_1 + f(P)area(D)_2 + …$$

$$=1.4 * 8.6 + 0.7 * 8.6 + 0.9 * 8.5 + 0.9 * 9.1 + 1.1 * 8.5 + 1.1 * 9.1= 53.26$$

15.3 Triple Integrals

Similar to double integrals except there is another degree of restriction
$$B=[a,b] \times [c,d] \times [p,q]$$
such that all points in $\mathbb{R}^3$
$$a \le x \le b, \quad c \le y \le d, \quad p \le z \le q$$

To integrate over this box we subdivide the box into subboxes
$$B_{ijk}=[x_{i-1},x_i] \times [y_{j-1},y_j] \times [z_{k-1}, z_k]$$
by choosing partitions of the three intervals
$$a= x_0 < x_1 < … < x_N = b$$
$$c = y_0 < y_1 < … < y_M = d$$
$$p = z_0 < z_1 < … < z_L = q$$

Here, N, M, and L are positive integers. The volume of $B_{ijk}$ is $\Delta V_{ijk}=\Delta x_i, \Delta y_j, \Delta z_k$ where
$$\Delta x_i = x_i - x_{i-1}, \quad \Delta y_j = y_j - y_{j-1}, \quad \Delta z_k = z_k - z_{k-1}$$
Then we choose a sample point $P_{ijk}$ in each sub box $B_{ijk}$ and form the Riemann sum…
$$S_{N,M,L}=\sum^N_{i=1}\sum^M_{j=1}\sum^L_{k=1} f(P_{ijk}) \Delta V_{ijk}$$
We write $P={{x_i},{y_j},{z_k}}$ for the partition and let $||P||$ be the maximum of the widths $\Delta x_i, \Delta y_j, \Delta z_k$. If the sums $S_{N,M,L}$ approach a limit as $||P|| \rightarrow 0$ for arbitrary choices of sample points, we say that $f$ is integrable over $B$. The limit value is denoted
$$\iiint_B f(x,y,z)dV = \lim_{||P|| \rightarrow 0} S_{N,M,L}$$

Theorem 1: Fubini’s Theorem for Triple Integrals

The triple integral of a continuous function $f(x,y,z)$ over a box $B= [a,b] \times [c,d] \times [p, q]$ is equal to the iterated integral
$$\iiint_B f(x,y,z)dV= \int^b_{x=a}\int^d_{y=c}\int^q_{z=p}f(x,y,z)dz \space dy \space dx$$
Furthermore, the iterated integral can be evaluated in any order

Example :: Triple Integral:

Evaluate
$$\iiint_B (xz + yz^2) dV \text{ for box: } 0 \le x \le 4, \quad 2 \le y \le 4, \quad 0 \le z \le 4$$

1. Rewrite:
   $$\int^4_0 \int^4_2 \int^4_0 (xz + yz^2)dzdydx$$
2. Evaluate the inner integral
   $$\int^4_0(xz + yz^2)dz=\frac{1}{2}xz^2 + \frac{1}{3}yz^3 \vert^4_0=\frac{1}{2}x(4)^2 + \frac{1}{3}y(4)^3-(0)= 8x + \frac{64}{3}y$$
3. Evaluate the middle integral
   $$\int^4_2(8x + \frac{64}{3}y)dy = 8x + \frac{32}{3}y^2 \vert^4_2 = 8x(4) + \frac{32}{3}(4)^2 - \left( 8x(2) \frac{32}{3}(2)^2 \right) = 16x + 128$$
4. Evaluate the outer integral
   $$\int^4_0 (16x + 128)dx = 8x^2 + 128x \vert^4_0 = 640$$

Theorem 2

The triple integral of a continuous function $f$ over the region
$$W \quad : \quad (x,y) \in D, \quad z_1(x,y) \le z \le z_2 \space (x,y)$$
is equal to the iterated integral
$$\iiint_W f(x,y,z) \space dV = \iint_D \left( \int^{z_2(x,y)}_{z=z_1(x,y)} f(x,y,z)dz\right)dA$$
Note:
More generally, integrals of functions of $n$ variables (for any $n$ ) arise naturally in many different contexts. For example, the average distance between two points in a ball in $R^3$ is expressed as a six-fold integral because we integrate over all possible coordinates of the two points. Each point has three coordinates for a total of six variables.

The volume of a region $W$ is defined as
$$V=\iiint_W 1dV$$

Example :: Integrating inner integral with equations:

Integrate $f(x,y,z)=x$ over the region $W$ in the first octant above $z=y^2$ and below $z=80-5x^2 -4y^2$
Rewrite:
$$\iint_D \int^{80-5x^2-4y^2}_{z=y^2}x$$

1. Now we find the ranges for the next two integrals. Knowing that the first quadrant starts at $x=0$ and $y=0$ we can assume those are the bottom constraints. For the top constants we can set the two equations equal to one another
   $$80-5x^2-4y^2 = y^2, \quad 16 - x^2 = y^2, \quad x^2 + y^2 = 16, \quad x = 4, y = \sqrt{16-x^2}$$
2. Now we rewrite the integral
   $$\int_0^4 \int_0^{\sqrt{16-x^2}} \int^{80-5x^2-4y^2}_{z=y^2}x$$
3. Calculate:
4. $$\int_0^4 \int_0^{\sqrt{16-x^2}} \int^{80-5x^2-4y^2}_{z=y^2}x = \frac{2048}{3}$$

Find the volume of the solid V in the first octant bounded by $x + y + z = 2$ and $x + y + 3z = 2$ NOTICE: REGION IS Z-SIMPLE

1. Set the equations to each other to find the bounds
   $$z:\quad z=2-x-y, \frac{2-x-y}{3} \quad \frac{2-x-y}{3}\le z \le 2-x-y$$
2. Knowing that the integral is z-simple we can solve that as the first integral of our problem
3. $$V = \iiint_W dV=\iint_D \int^{2-x-y}_{z=\frac{2-x-y}{3}}dz \space dA$$
4. Solve for the y-variable in this equation by setting the equations equal to one another
   $$2 - x - y = \frac{2-x-y}{3}, \quad y=2-x$$
5. Plug in the new constraints
6. $$\int^b_a \int^{2-x}<em>0 \int^{2-x-y}</em>{z=\frac{2-x-y}{3}}dz \space dy \space dx$$
7. Find the final constraints for x
8. $$0 \le x \le b, \quad 0 \le y \le 2-x, \quad 2-x-y \le z \le \frac{2-x-y}{3}$$
9. setting $x=0$ in the y-variable constraint, we can confirm that the top constraint is $x=2$
10. $$0 \le x \le 2, \quad 0 \le y \le 2-x, \quad 2-x-y \le z \le \frac{2-x-y}{3}$$
11. Plug in
12. $$\int^2_0 \int^{2-x}0 \int^{2-x-y}_{z=\frac{2-x-y}{3}}dz \space dy \space dx$$
13. Solve
14. $$\int^2_0 \int^{2-x}0 \int^{2-x-y}_{z=\frac{2-x-y}{3}}dz \space dy \space dx=\frac{8}{9}$$

Let $W$ be the region bounded by $z=4-y^2, y=2x^2$ and the plane $z=0$. Calculate the volume of $W$ as a triple integral

1. Find out which dimension the equation is simple for: I think the volume would be z simple since we know z is constrained between $z=0$ and $z=4-y^2$.
   $$\iiint_W dV = \iint_D \int_0^{4-y^2} dz \space dA$$
2. Then we can solve for y by plugging into the equation with z and y
   $$(0) = 4-y^2, \quad y = 2$$
   $$\int^b_a \int_{2x^2}^{2} \int_0^{4-y^2} dz \space dy \space dx$$
3. Find constraints for x: We can set y = 2 and say $x = 1, -1$ with the 2nd equation.
   $$\int^1_{-1} \int_{2x^2}^{2} \int_0^{4-y^2} dz \space dy \space dx = \frac{128}{21}$$

15.4 Polar Coordinates

Rectangular Coordinates

The typical coordinate system relying on an absolute origin

Polar Coordinates

Labeling a point $P$ through coordinate ordering of $(r, \theta)$ where $r$ is the distance to the origin $O$ and $\theta$ is the angle between $\overline{OP}$. $\theta$ moves counterclockwise

Example :: Rectangular to Polar:

Convert Rectangular to Polar Coordinates

1. (0,-1)
   $$r = \sqrt{0^2 + (-1)^2})=1 \quad \tan^{-1}(1)= \frac{\pi}{4}$$
2. (3, $\sqrt{3}$)
   $$r = \sqrt{3^2 + (\sqrt{3})^2}= \sqrt{12} \quad \tan^{-1}(\sqrt{12})=1.2898$$
3. $(-1, \sqrt{3})$
   $$r=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{10} \quad \tan^{-1}(\sqrt{10}) = 1.26451$$

Example :: Integrating a circle:

Assume that $a=1$. Integrate $f(x,y)=y(x^2 + y^2)^3$ over $D$ using polar coordinates.
$$\iint_D y(x^2 + y^2)^3 dA$$
$$y=a \sin \theta, \quad x = a \cos \theta$$

$$\iint_D (a \sin \theta )((a \cos \theta )^2 + (a \sin \theta )^2) dA= \int^\pi_{\theta = 0} \int^a_0 (a \sin \theta )(a^2)^3 dA$$ $$\int^\pi_{\theta = 0} \int^1_0 ( \sin \theta )(a^8) da \space d\theta = \frac{2}{9}$$

Evaluate by changing to polar coordinates

$$\int^{\pi/2}_0 \int^{9}_0 \sqrt{r^2\cos^2\theta + r^2\sin^2\theta} \space da \space d \theta$$

$$\int^{\pi/2}_0 \int^2_0 6 \cdot r^3\cos\theta \cdot \sin\theta drd\theta$$

$$\int_0^{\pi/2} \int^{\sqrt{\sin2 \theta}}_0 r^3 \cos \theta \sqrt{17}drd\theta$$

15.4.2 Cylindrical and Spherical Coordinates

Example :: Converting Cylindrical to Rectangular Coordinates:

Find the rectangular coordinates of the point $P$ with the cylindrical coordinates $(r,\theta,z)=(2, \frac{3\pi}{4}, 5)$

1. $$x=r\cos\theta = 2\cos \frac{3\pi}{4}=2 \left(-\frac{\sqrt{2}}{2}\right)=-\sqrt{2}$$
   $$y=r\sin\theta = 2\sin \frac{3\pi}{4}=2 \left(\frac{\sqrt{2}}{2}\right)=\sqrt{2}$$
2. $$z=z=5, \text{ANS: } (-\sqrt{2}, \sqrt{2}, 5)$$
3. Convert from cylindrical to rectangular coordinates $(6, \frac{\pi}{3}, -8)$
4. $$x = r\cos \theta = 6 \cos \frac{\pi}{3} = 6(\frac{1}{2})=3$$
5. $$x = r\sin \theta = 6 \sin \frac{\pi}{3} = 6(\frac{\sqrt{3}}{2})=3\sqrt{3}$$
6. Convert from rectangular to cylindrical coordinates $(12, 4\sqrt{3}, 10)$
7. $$r = \sqrt{x^2+y^2} = \sqrt{12^2+(4\sqrt{3})^2} = 8\sqrt{3}$$
8. $$\theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{4\sqrt{3}}{12}=\tan^{-1} \frac{\sqrt{3}}{3}=\frac{\pi}{6}$$

Example :: Use cylindrical coordinates to integrate:

Use cylindrical coordinates to calculate the triple integral $\iiint_W f(x,y,z)dV$ for the function $f(x,y,z)=\frac{1}{2116}z\sqrt{x^2+y^2}$ and the region $x^2 + y^2 \le z \le 46 - (x^2 + y^2)$

1. Transform rectangular into cylindrical coordinates
   $$f(x,y,z)=\frac{1}{2116}z\sqrt{x^2+y^2}=\frac{1}{2116}z(r)$$
   $$x^2 + y^2 \le z \le 46 - (x^2 + y^2) = (r^2)\le z \le 46 - (r^2)$$
2. Solve for the values to integrate
   $$r^2 = 46 - r^2, \quad 2r^2 = 46, \quad r = \sqrt{46/2}=\sqrt{23}$$
3. Plug into the new coordinates
4. $$\int_0^{2\pi} \int_0^{\sqrt{23}} \int_{r^2}^{46-r^2} \frac{1}{2116}zr \cdot r \space dzdrd\theta = \frac{46 \pi \sqrt{23}}{15}$$
5. Find the volume $V$ of the region appearing between the two surfaces $z=x^2 + y^2$ and $z=128 - x^2 - y^2$
6. Transform items into spherical terms
   $$z=x^2 + y^2 = r^2 \quad z = 128 - (x^2 + y^2) = 128- r^2$$
7. Solve for radius
8. $$r^2 = 128 - r^2, \quad 2r^2 = 128, \quad r = 8$$
9. Set up integral:
10. $$\int^{2\pi}<em>0 \int^8_0 \int</em>{r^2}^{128-r^2} r \cdot dzdrd\theta = 4096\pi$$

Spherical Coordinates

Spherical coordinates make use of the fact that a point $P$ on a sphere of radius $\rho$ is determined by two angular coordinates $\theta$ and $\phi$

Example :: Integrating using cylindrical coordinates:

Use cylindrical coordinates to calculate $\iiint_W f(x,y,z)dV$ for $f(x,y,z)=x$ while $x^2 + y^2 \le 9$ and $x \ge 0$ and $y \ge 0$ and $-4 \le z \le 4$

1. Solve for $\rho$
2. $$x^2 + y^2 = 9,\quad \rho^2 = 9, \quad \rho = 3$$
3. $$\int^{\theta_2}<em>{\theta_1} \int^{r_2(\theta)}</em>{r_1(\theta)} \int^{z_2(r,\theta)}<em>{z_1(r, \theta)} x \space dzdrd \theta = \int^{\theta_2}</em>{\theta_1} \int^{r_2(\theta)}<em>{r_1(\theta)} \int^{z_2(r,\theta)}</em>{z_1(r, \theta)} r\cos \theta \cdot r \space dzdrd \theta$$
4. $$\int^{2\pi}<em>{0} \int^{3}</em>{0} \int^{4}_{-4} r\cos \theta \cdot r \space dzdrd \theta$$

15.5 Applications of Multiple Integrals

$$\text{Radius of gyration: }r_g = \left(\frac{I}{M}\right)^{1/2}$$
Random variables $X$ and $Y$ have joint probabliity density function $p(x,y)$ if…
$$P(a \le X \le b; c \le Y \le d) = \int^b_{x=a} \int^d_{y=c} p(x,y) dydx$$
A joint probability density function must satisfy $p(x,y) \ge 0$ and
$$\int^{\infty}<em>{x=-\infty}\int^{\infty}</em>{y=-\infty} p(x,y)dydx = 1$$

Examples :: Density:

Find the total mass of the rectangle $0 \le x \le 4, 1 \le y \le 7$ assuming a mass density of $\delta (x,y) = 4x^2 + y^2$

1. Use the formula for average $\overline{f}$
   • $$\overline{f}=\frac{\iint_D f(x,y)dA}{\iint_D 1dA}$$
2. Find the bounds
   • $$\int^{12}<em>{x=6} \int^{\sqrt{x-6}}</em>{-\sqrt{x-6}} 1dxdy = 8\sqrt{6}$$