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# System prepended metadata

title: Week 12 (Part 2) & Week 13

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During next 3 weeks, we plan to discuss the following key concepts

1. Integration of functions of several variables: multiple integrals
2. Change of variables
3. Applications of multiple integrals

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## 4.1 Multiple Integrals over rectangular regions

The **multiple integral** is a *definite integral* of a function of several variables $f(x_1,\dots,x_n)$. Integrals of a function of two variables over a region in $\mathbb R^2$ are called *double integrals*, and integrals of a function of three variables over a region of $\mathbb R^3$ are called *triple integrals*.

Let us recall the definition of a definite integral for a continuous function $f:[a,b] \to \mathbb R$. 

:::info

Divide $[a,b]$ into $n$ subintervals $[x_{i-1},x_i],\, i=1,\dots, n$ of equal length given by $\Delta x = (b-a)/n$ with $x_0=a$, $x_i=x_0+i\Delta x$, and $x_n=b$. 

Let $x_i^* \in [x_{i-1},x_i]\,, i=1,\dots, n$ be sample points. Define $$\int_a^b f(x)\, dx := \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*)\, \Delta x\,.$$ Since $f$ is continuous on $[a,b]$ the limit on the RHS is independent of the choice of $\{x_i^*\}_{i=1}^n$ and hence, $\int_a^b f(x)\, dx$ is well-defined.

![image](https://hackmd.io/_uploads/HkLntiH7ye.png)

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This generalises to higher dimensions. In particular, in two dimensions we can visualise this as follows.

![image](https://hackmd.io/_uploads/BkKnxnrXye.png)  $$\int_{c}^d\int_{a}^b f(x,y)\,dx\,dy = \lim_{m,n \to \infty} \sum_{j=1}^m\sum_{i=1}^n f(x_{ij}, y_{ij})\Delta x\Delta y $$![image](https://hackmd.io/_uploads/HyGWbnSQ1x.png)
<br>

:::info

**Defintion** (Multiple Integrals over Rectangles)

In general, if $D\subset \mathbb R^n$ is a *rectangular* region given by $R=[a_1,b_1]\times \dots \times [a_n,b_n]$, and $f: D \to \mathbb R$  is a continuous function, then the integral of $f$ over $R$ denoted by $$\int \cdots \int_R\, f(x_1,\dots,x_n)\, dx_1 \cdots dx_n$$ can be defined as a limit of Riemann sums  $$\lim_{m_1,\dots,m_n \to \infty}\sum_{i_n=1}^{m_n} \cdots \sum_{i_1=1}^{m_1} f( y^*_{i_1,\dots,i_m})\Delta x_1\cdots \Delta x_n $$ where $y^*_{i_1,\dots,i_m} \in \mathbb R^n$ is the sample point taken from the $(i_1,\dots,i_n)-$*rectangle* which is constructed by dividing each $[a_j,b_j]$ into $m_j$ subintervals of equal length of $$\Delta x_j = (b_j-a_j)/m_j\,.$$

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But this definition is not useful in actually evaluating integrals! So, let us try compute some multiple integrals.

### 4.1.1 Iterated Integrals


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In order to compute multiple integrals, we proceed as follows:
1. We keep $x_2,\dots,x_n$ fixed and evaluate the definite integral with respect to $x_1$. This is called *partial integration with respect to $x_1$*.  The result is a function of $x_2,\dots,x_n$ independent of $x_1$. 
2. We repeat the process for each variable $x_2, \dots, x_n$ iteratively, each time partially integrating with respect a variable in the given order. 

This is called an **iterated integral**:
$$\int \cdots \int_R\, f(x_1,\dots,x_n)\, dx_1 \cdots dx_n = \int_{a_n}^{b_n}\left[\cdots\left[\int_{a_1}^{b_1}f(x_1,\dots,x_n)\, dx_1\right]\cdots\, \right]dx_n$$
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- Note that the iterated integral can be written in many orders (e.g., $x_n$ first, $x_{n-1}$ second, and so on). We will see later that if $f$ continuous, this order will not matter. 
- In two dimensions we can easily visualise this.![image](https://hackmd.io/_uploads/HJ3Vwnr7kg.png =500x350)
Write $$A(x) = \int_c^d f(x,y)\, dy$$ and then $$\iint_R f(x,y) \,dy\, dx= \int_a^b A(x)\, dx = \int_a^b \left[\int_c^d f(x,y) \,dy\right]\, dx\,.$$


**Example 1**

Compute the following iterated integral. $$ \int_0^{\pi/2} \int_{0}^\pi \sin(x+y)\,dx\,dy.$$

:::spoiler Answer

First integrating with respect to $x$ (treating $y$ as a constant)
$$\int_{0}^\pi \sin(x+y)\,dx = -\cos (x+y)\Big|_0^\pi=-(\cos(\pi+y)-\cos(y))=2\cos(y)$$This gives us the inner integral. Then $$\int_0^{\pi/2} \int_{0}^\pi \sin(x+y)\,dx\,dy=2\int_{0}^{\pi/2}\cos(y)\,dy = \sin(y)\Big|_0^{\pi/2}=1.$$
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<br>


**Example 2**

Compute the following: $$ \int_1^2 \int_{0}^3 x^2y\,dy\,dx\,\,,\,\,\,\int_{0}^3 \int_1^2 x^2y\,dx\,dy\,\,,\,\,\,\left(\int_{1}^2 x^2\, dx\right)\left(\int_0^3 y\,dy\right)$$ Are they the same?


:::spoiler Answer
$$\int_1^2 \int_{0}^3 x^2y\,dy\,dx = \int_1^2 x^2y^2\Big|_{y=0}^{y=3} \,dx = \int_1^2 9x^2 \,dx = 3x^3\Big|_{y=1}^{y=2} =21\,. $$

$$\int_0^3 \int_{1}^2 x^2y\,dx\,dy = \int_0^3 \frac{1}{3}x^3y^2\Big|_{x=1}^{x=2} \,dy = \int_1^2 \frac{7}{3}y^2  \,dy = \frac{7}{9}y^3\Big|_{y=0}^{y=3} =21\,.$$

$$\left(\int_{1}^2 x^2\, dx\right)\left(\int_0^3 y\,dy\right) = \frac{x^3}{3}\Big|_{x=1}^{x=2} \times \frac{y^2}{2}\Big|_{y=0}^{y=3} = \frac{7}{3} \times \frac{9}{2} = 21\,.$$

All of them have the same value.
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<br>

**Example 3**

Compute the following two iterated integrals. $$ \int_{0}^\pi\int_{1}^2 y \sin (xy)\,dx\,dy\,\,,\,\,\,\int_1^2 \int_{0}^\pi y \sin (xy) \,dy\,dx\,.$$Are they the same?
<br>

:::spoiler Answer
Note that \begin{align}\int_{0}^\pi\int_{1}^2 y \sin (xy)\,dx\,dy = \int_0^\pi -\cos(xy)\Big|_{x=1}^{x=2}\,dy &= \int_0^\pi [\cos(2y)-\cos(y)]\,dy \\ &=\left.\frac{\sin (2y)}{2} - \sin(y)\right]_{y=0}^{\pi}= 0\,.\end{align}

However, the second iterated integral is much more difficult to evaluate. By applying integration by parts,
$$\int_{0}^\pi y \sin (xy) \,dy = \frac{-\pi x \cos(\pi x) + \sin(\pi x)}{x^2}\,.$$So,
$$\int_1^2 \int_{0}^\pi y \sin (xy) \,dy\,dx = \int_1^2 \frac{-\pi x \cos(\pi x) + \sin(\pi x)}{x^2}\,\,dx = -\frac{\sin(\pi x)}{x}\Big|_1^2=0.$$

Both integrals evaluate to $0$.
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<br>

### 4.1.2 Fubini's Theorem

In the last two examples, we noted that the order of integration did not change the outcome. This is true in general.

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**Theorem (Fubini)**

if $D\subset \mathbb R^n$ is a *rectangular* region given by $[a_1,b_1]\times \dots \times [a_n,b_n]$, and $f: D \to \mathbb R$  is a continuous function, then for any permutation/reaarangement $\sigma$ of $\{1,\dots,n\}$, $$\int_{a_n}^{b_n}\cdots\int_{a_1}^{b_1}f\,\,\, dx_1 \cdots dx_n = \int_{a_{\sigma(n)}}^{b_{\sigma(n)}}\cdots\int_{a_{\sigma(1)}}^{b_{\sigma(1)}}f\,\,\, dx_{\sigma(1)} \cdots dx_{\sigma(n)}\,.$$
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<br>


In particular, we have that $$\int_{a}^b\int_{c}^d f(x,y)\,dy\,dx = \int_{c}^d\int_{a}^b f(x,y)\,dx\,dy\,.$$ provided $f$ is contnuous. So, in **Example 3**, we could have easily concluded that the value of the second integral without computing it. 
<br>

Also, in the special case of $f(x,y)=g(x)h(y)$ (separable), we can write
\begin{align}
\int_{a}^b\int_{c}^d f(x,y)\,dy\,dx &= \int_{a}^b\int_{c}^d g(x)h(y)\,dy\,dx \\&=  \int_{a}^b g(x) \left(\int_{c}^d h(y)\,dy\right)\,dx = \left(\int_{c}^d h(y)\,dy\right)\left(\int_{a}^b g(x)\,dx\right)
\end{align}

We observed a special case of this in **Example 2**. This holds in higher dimensions as well.
<br>

**Example 4** 

Find $$\iiint_V e^{x+2y+3z}\,dx\,dy\,dz$$ where $V=[0,6]\times[0,3]\times[0,2]$. 
<br>

:::spoiler Answer

Note that the integrand is continuous and seperable. So, we have that 
\begin{align}
\iiint_V e^{x+2y+3z}\,dx\,dy\,dz &= \int_0^6 e^x\,dx \int_0^3 e^{2y}\,dy \int_{0}^2 e^{3z}\,dz\\ &= e^{x}\Big|_{0}^6 \times \frac{1}{2}e^{2y} \Big|_{0}^3 \times \frac{1}{3}e^{3z}\Big|_{0}^2 =\frac{1}{6}(e^6-1)^3 \,.
\end{align}
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<br>

**Example 5** 

Find $$\iiint_V zy\sin(xz)\,dx\,dy\,dz$$ where $V=[1,2]\times[0,\pi/2]\times[0,\pi]$. 
<br>

:::spoiler Answer

Note that the integrand is continuous, and *partially* seperable. So, we have that 
\begin{align}
\iiint_V zy\sin(xz)\,dx\,dy\,dz &= \int_0^{\pi/2} y\,dy \int_0^{\pi}\int_1^2 z \sin(xz)\,dx\,dz = 0\,.
\end{align}because from Example 2, the second integral is $0$ (write $z$ instead of $y$ in Example 2).
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<br>

### 4.1.3 Volumes under graphs 

From the defintion of multiple integrals it is clear that if $f \geq 0$ on a rectangle $R$, then the double integral $\iint_R f(x,y)\, dx\,dy$ gives the volume of the solid under the graph of $f(x,y)$ and over the rectangle $R$. 

![image](https://hackmd.io/_uploads/HkRN_yo7yg.png =275x260)![image](https://hackmd.io/_uploads/ryB3XJi7yx.png)

Let's use this idea to compute volumes.

**Example 6** 

Find the volume of the solid bounded by the elliptic paraboloid $x^2+2y^2+z=16$, the planes $x=2$ and $y=2$, and the three coordinate planes. 

![image](https://hackmd.io/_uploads/H1HEWkiQJl.png)

:::spoiler Answer

The solid is under the graph $z=16-x^2-2y^2$ over the rectangle $[0,2]\times[0,2]$. So the volume of the solid is 
\begin{align}
\int_0^2\int_0^2 (16x-x^2-2y^2)\,dx\,dy &= \int_0^2 \left.16x-\frac{x^3}{3}-2y^2x\right]_{x=0}^{x=2}\,dy \\&=\int_0^2 \left(\frac{88}{3} -4y^2 \right)\,dy = \left.\frac{88y}{3}-\frac{4y^3}{3}\right]_{y=0}^{y=2} = 48\,
\end{align}

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<br>


---
**Reference** 
1. *Chapter 15.1 & 15.6* : Stewart, J. (2012). Calculus (8th ed.). Boston: Cengage Learning. 
2. *Chapter 14.1 & 14.3* : Strang G., Calculus (3rd ed.). [https://ocw.mit.edu/](https://ocw.mit.edu/courses/res-18-001-calculus-fall-2023/pages/textbook/) 
3. *Chapter 3.1 & 3.3* : Corral, M. (2021). Vector Calculus. [https://www.mecmath.net/](https://www.mecmath.net/) 

---
<br>


## 4.2 Multiple integrals over general regions

Since regions in $\mathbb R^n$ are not necessarily rectangular, we would like to extend the concept of integral to more general regions. 

Suppose $D$ is bounded region in $\mathbb R^n$, i.e., there is a rectangle $R = [a_1,b_1]\times \dots \times [a_n,b_n]$ in $\mathbb R^n$ such that $D \subseteq R$\,, and $f:D \to \mathbb R$ is continuous. 

![image](https://hackmd.io/_uploads/S13oWlsmye.png)

Then, we can define a new function $F:R\to \mathbb R$ as follows:$$F =\begin{cases} f & \text{on}\,\,\, D \\ 0 & \text{on}\,\,\,R\setminus D\end{cases}$$

![image](https://hackmd.io/_uploads/BJ0bGeo7kx.png)

Now, we can define the integral of $f$ over $D$ as follows:
:::info 

**Defintion** (Multiple Integral)

$$\int\cdots\int_{D} f\,\,\, dx_1\,\cdots\,dx_n := \int\cdots\int_{R} F\,\,\, dx_1\,\cdots\,dx_n$$where the RHS above is defined as the limit of Riemann sums as in Section 3.8. Even though $F$ is not continuous on $R$, the limit of Riemann sum exists, is idependent of the choice of sample points as long as the boundary of $D$ is *reasonably nice*, and is also independent of the choice of $R$. 
:::

### 4.2.1 Double Integrals

Let's look at some *reasonably nice* regions in $\mathbb R^2$. 
- **Type I regions**: $D=\{(x,y)|\, a \leq x \leq b\,,\,\,g_1(x)\leq y \leq g_2(x)\,\}$ where $g_1$ and $g_2$ are continuous functions.
![image](https://hackmd.io/_uploads/S142QloQ1g.png)
In this case, $$\iint_D f(x,y)\, dx\,dy = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx\,.$$

- **Type II regions**: $D=\{(x,y)|\, c \leq y \leq d\,,\,\,h_1(y)\leq x \leq h_2(y)\,\}$ where $h_1$ and $h_2$ are continuous functions.
![image](https://hackmd.io/_uploads/HJsx4es7ke.png)      ![image](https://hackmd.io/_uploads/SyiHNxo7Jx.png)
In this case, $$\iint_D f(x,y)\, dx\,dy = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy\,.$$

- There are more general *resoanbly nice* regions in $\mathbb R^2$. We will discuss them later. ![image](https://hackmd.io/_uploads/SkHmRxi7Je.png)![image](https://hackmd.io/_uploads/HyZKJ-iQyx.png)![image](https://hackmd.io/_uploads/Bk1ZgWsQkx.png)





**Example 1**

Let $D$ be the region bounded by $y=2x^2$ and $y=1+x^2$. Evaluate$$\iint_{D} 1\,\,dx\,dy\,.$$

:::spoiler Answer

We note that the two parabolas intersect: $2x^2 = 1+x^2$ if and only if $x=\pm 1$. 

![image](https://hackmd.io/_uploads/HkvUtgoXkl.png)

\begin{align}
\iint_{D} 1\,\,dx\,dy = \int_{-1}^{1}\int_{2x^2}^{1+x^2}1\,\,dy\,dx = \int_{-1}^{1} y\Big|_{y=2x^2}^{y=1+x^2}\,dx = \int_{-1}^1 (1-x^2) \,dx = x-\frac{x^3}{3}\Big|_{x=-1}^{x=1} =\frac{4}{3}. 
\end{align}
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<br>

What we computed is, in fact, the area bounded by the two parabolas. In general, the area of a region $D$ in $\mathbb R^2$ is *numerically* equal to the volume of the cylinder of height $1$ with $D$ as the cross section (equivalently, to the volume under the graph of constant function $1$ over the region $D$.) 
![image](https://hackmd.io/_uploads/HJ4WAh2Q1g.png)



:::info
 $$ \text{Area}\,(D) = \iint_D 1\,\, dx\,dy$$
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<br>



**Example 2**

Let $D$ be the region bounded by the curves $y=x-1$ and $y^2=2x+6$. Find $$\iint_D xy\, \,dx\,dy$$

:::spoiler Answer

We note that the two curves intersect at points where $(x-1)^2=2x+6$. This gives us that $x^2-4x+5=0$ which gives us $x=5,-1$. So, there are two points of intersection $(5,4)$ and $(-1,-2)$.

![image](https://hackmd.io/_uploads/B17gGWjQke.png)

It is easier to think of the region of integration as a type II one.

Then,\begin{align}
\iint_D xy\, \,dx\,dy = \int_{-2}^4 \int_{y^2/2 -3}^{y+1} xy\,\,dx\,dy &= \int_{-2}^4 \frac{x^2y}{2}\Big|_{x=y^2/2 -3}^{x=y+1}\,\,dy\\ &= \frac{1}{2}\int_{-2}^4 y(y+1)^2 - y(y^2/2-3)^2\,dy=36\,.
\end{align}
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<br>


**Example 3**

Find $$\int_0^1 \int_{x}^1 2\sin(y^2)\,\,dy\,dx\,.$$

:::spoiler Answer
Note that $\sin(y^2)$ has no closed-form antiderivative. So, we change the order of integration.
![image](https://hackmd.io/_uploads/rkQafWjQkl.png)![image](https://hackmd.io/_uploads/HkfbQWoQ1g.png)
\begin{align}
\int_0^1 \int_{x}^1 2\sin(y^2)\,\,dy\,dx\,=\int_0^1\int_{0}^y 2\sin(y^2) \,dx\,dy &= \int_0^1 2\sin(y^2) x\Big|_{x=0}^y \, dy\\&= \int_0^1 2y\sin(y^2)\,dy\\ &= -\cos(y^2)\Big|_0^1 =  1 - \cos 1.
\end{align}
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<br>

**Example 4**

Find the volume of the solid that lies under the paraboloid $z=x^2+y^2$ and above the region in the $xy-$plane bounded by the curves $y=2x$ and $y=x^2$.
<br>


:::spoiler Answer

The two curves intersect when $2x=x^2$, i.e., $x=0,2$. So, the region $D$ on $xy-$plane is as follows.
![image](https://hackmd.io/_uploads/ryceV-sXJl.png)
Note that each point $(x,y)$ on $D$, $z=x^2+y^2$. Hence,
\begin{align}
\text{Volume}=\iint_D x^2+y^2\,dx\,dy &= \int_0^4\int_{x^2}^{2x} x^2+y^2 \,dy\,dx \\&= \int_0^2 x^2y+y^3/3\Big|_{y=x^2}^{y=2x} \, dx\\&=\int_0^2 (2x^3 + 8x^3/3 - x^4 -x^6/3 )  \,dx = 216/15
\end{align}
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<br>



### 4.2.2 Triple Integrals

Consider the following *reasonably nice* regions in $\mathbb R^3$.
- **Type I** 
![image](https://hackmd.io/_uploads/SkeFwTnX1g.png)
$$\iiint_E f\,\, dx\,dy\,dz = \iint_D \left[\int_{u_1(x,y)}^{u_2(x,y)} f\,dz\right]\,dx\,dy $$
- **Type II** 
![image](https://hackmd.io/_uploads/B1zfd6nXye.png)
$$\iiint_E f\,\, dx\,dy\,dz = \iint_D \left[\int_{u_1(y,z)}^{u_2(y,z)} f\,\,dx\right]\,dy\,dz $$
- **Type III**
![image](https://hackmd.io/_uploads/r1OBtp2myg.png)
$$\iiint_E f\,\, dx\,dy\,dz = \iint_D \left[\int_{
u_1(x,z)}^{u_2(x,z)} f\,\,dy\right]\,dx\,dz $$

In each case, once we evaluate the innermost integral, the integral is reduced to a double integral over $D$, which we know how to evaluate provided $D$ is *reasonably nice*.
<br>


**Example 5** Evaluate the integral $$\iiint_E 6z \,\,dx\,dy\,dz\,,$$ where $E$ is the solid tetrahedron bounded by the four planes $x=0, y=0, z=0,$ and $x+y+z=1$.
<br>

:::spoiler Answer

![image](https://hackmd.io/_uploads/HJbljT2Xkx.png)

Let $D$ be the triangle bounded by $x=0, y=0,$ and $x+y=1$,
![image](https://hackmd.io/_uploads/ByRzlCnXke.png)
then
\begin{align}\iiint_E 6z \,\,dx\,dy\,dz = \iint_D\int_{0}^{1-x-y} 6z \,\,dz\,dx\,dy &= \iint_D 3z^2\Big|_{z=0}^{z=1-x-y} \,dx\,dy \\&= \iint_D(1-x-y)^2\,dx\,dy \,.
\end{align}So,
\begin{align}\iiint_E 6z \,\,dx\,dy\,dz =\int_{0}^1\int_{0}^{1-x} 3(1-x-y)^2\,dy\,dx &= \int_0^1 -(1-x-y)^3\big|_{y=0}^{y=1-x}\,dx\\ &=\int_{0}^1(1-y)^3dy=\frac{1}{4}\,.\end{align}
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<br>

**Example 6** Evaluate the integral $$\iiint_E 1 \,\,dx\,dy\,dz\,,$$ where $E$ is the solid sphere $x^2+y^2+z^2 \le 1$.
<br>

:::spoiler Answer
\begin{align}
&\iiint_E 1 \,\,dx\,dy\,dz \\&= \int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} 1 \,\,dz\,dy\,dx\\ &= \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \sqrt{1-x^2-y^2}\,dy\,dx\\ &=\int_{-1}^1 \left.y\sqrt{1-x^2-y^2}+(1-x^2)\sin^{-1}\frac{y}{\sqrt{1-x^2}}\right]_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}}\,dx\\&=\pi\int_{-1}^{1} (1-x^2)\,dx = \frac{4\pi}{3}\,.\end{align}
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<br>

What we computed is, in fact, the volume of the solid sphere $E$. In general, we have the following.

:::info
$$\text{Volume}(E)=\iiint_E 1 \,\,dx\,dy\,dz$$
:::
<br>

**Example 7**

Find the volume of the solid bounded by the paraboloid $z=x^2+y^2$, and the planes $y=2x$, $y=x^2$ and $z=0$. 
<br>

:::spoiler Answer
Note that this solid region $E$ is the same as the region under the graph of  $z=x^2+y^2$ above the two dimensonal region $D$ bounded by $y=2x$ and $y=x^2$ (see Example 4).

$$\text{V}(E)=\iiint_E1\,dz\,dx\,dy = \iint_D\int_{0}^{x^2+y^2}1\,dz\,dx\,dy = \iint_Dz\Big|_{z=0}^{x^2+y^2}\,dx\,dy =\iint_D (x^2+y^2)\,\,dx\,dy \,.$$So, from Example 4, Volume$(E)=\frac{216}{35}\,.$
:::
<br>

**Example 8**

Change the order of integration of
$$\int_0^1\int_{0}^{x^2}\int_0^y f(x,y,z)\,dz\,dy\,dx$$ to  integrating first with respect to $x$, then $z$, and finally, with respect to $y$.<br> 

:::spoiler Answer
Let $E$ be the region of integration. $E$ is bounded by the planes $z=0$, $z=y$, $y=x^2$ and $x=1$. The projection of $E$ to $xy-$plane is the region bounded by $y=0$ and $y=x^2$ and $x=1$ (See $D_1$ below). 
$$\int_0^1\int_{0}^{x^2}\int_0^y f(x,y,z)\,dz\,dy\,dx = \iint_{D_1}\int_{0}^{y} f(x,y,z)\,dz\,dy\,dx\,.$$

![image](https://hackmd.io/_uploads/B1znz16mye.png =225x170)![image](https://hackmd.io/_uploads/rkE-xBJ4kx.png =225x170)![image](https://hackmd.io/_uploads/Skq18yTXyx.png =224x170)

The projection of $E$ to the $zy-$plane is the region $D_2$ below bounded by $z=y$, $z=0$ and $y=1$ and above $D_2$, $x$ varies from $x=\sqrt{y}$ to $x=1$. So, $$\int_0^1\int_{0}^{x^2}\int_0^y f(x,y,z)\,dz\,dy\,dx = \int_0^1\int_{0}^{y}\int_{\sqrt{y}}^1 f(x,y,z)\,dx\,dz\,dy$$
:::
<br>

### 4.2.3 Properties of Integrals

Similar to one dimensional case, the multiple integral has the following propert. All of them follow directly from the definition of the integral as a limit of Riemann sums. 

:::success

**Theorem**

Suppose $D \subseteq \mathbb R^n$ has a *nice* boundary. Let $f,g : D \to \mathbb R$ be continuous, and $k$ be a constant. Then,
- $$\idotsint_D (f+g)\,dx_1\,\cdots\,dx_n = \idotsint_D f\,dx_1\,\cdots\,dx_n + \idotsint_D g\, dx_1\,\cdots\,dx_n$$
- $$\int\cdots\int_{D} k\,f\,dx_1\,\cdots\,dx_n=k \int\cdots\int_{D} f\, dx_1\,\cdots\,dx_n\,.$$
- If $D_1$ and $D_2$ are nice regions such that $D_1\cup D_2=D$ and $D_1\cap D_2$ has dimension less than $n$, then
$$\int\cdots\int_{D} f\,\,\, dx_1\,\cdots\,dx_n = \int\cdots\int_{D_1} f\,\,\, dx_1\,\cdots\,dx_n + \int\cdots\int_{D_2} f\,\,\, dx_1\,\cdots\,dx_n\,.$$
- If $f \leq g$ on $D$, then
$$\int\cdots\int_{D} f\,\,\, dx_1\,\cdots\,dx_n \leq \int\cdots\int_{D} g\,\,\, dx_1\,\cdots\,dx_n$$
- If $m \leq f \leq M$ on $D$, then
$$m\cdot \text{Volume}(D)\leq \int\cdots\int_{D} f\,\,\, dx_1\,\cdots\,dx_n \leq M\cdot \text{Volume}(D)\,.$$
:::
<br>

**Example 9**

Estimate the integral $$\iint_D e^{\sin x \cos y} \,dx\,dy$$where $D$ is the disc of radius $2$ centred at the origin.
<br> 

:::spoiler Answer

Note that $-1\leq \sin x \cos y \leq 1$. So, $e^{-1} \leq e^{\sin x \cos y} \leq e$. Also, the area of D is $4\pi$. So, from the last property of integral, the following *crude* estimate holds.
$$4\pi e^{-1} \leq \iint_D e^{\sin x \cos y} \,dx\,dy \leq 4\pi e$$
:::
<br>

**Example 10**

Evaluate $$\iint_D y \, dx\,dy$$
where $D$ is the region below.
![image](https://hackmd.io/_uploads/SyCUuCAmJx.png)
<br>

:::spoiler Answer

We divide the region $D=D_1 \cup D_2$ where $D_1$ is the part above the $x-$axis and the $D_2$ is the part below the $x-$axis. Then $D_1$ and $D_2$ are both type II regions.
<br>

\begin{align}
\iint_D y \, dx\,dy
&=\iint_{D_1} y \, dx\,dy +\iint_{D_2} y\, dx\,dy\\ 
&=\int_0^1\int_{\sqrt{y}-1}^{y-y^3} y\,dx \,dy + \int_{-1}^0\int_{-1}^{y-y^3} y\,dx \,dy \\ 
&=\int_{0}^1 yx\Big|_{x=\sqrt{y}-1}^{x=y-y^3} \,dy + \int_{-1}^0 yx\Big|_{x=-1}^{x=y-y^3} \,dy \\ &=\int_{0}^1 y(y-y^{3}-\sqrt{y}+1)\,dy+\int_{-1}^0 y(y-y^3+1) \,dy \\&= -\frac{2}{15}\,.
\end{align}

:::
<br>

**Example 11**

Calculate the area of the region $D$ in Example 10.
<br>

:::spoiler Answer
\begin{align}
\text{Area} (D)
&=\iint_{D_1} 1 \, dx\,dy + \iint_{D_2} 1 \, dx\,dy\\ 
&=\int_0^1\int_{\sqrt{y}-1}^{y-y^3} 1\,dx \,dy +\int_{-1}^0\int_{-1}^{y-y^3} 1\,dx \,dy \\ 
&=\int_{0}^1 x\Big|_{x=\sqrt{y}-1}^{x=y-y^3} \,dy + \int_{-1}^0 x\Big|_{x=-1}^{x=y-y^3} \,dy \\ &=\int_{0}^1 (y-y^{3}-\sqrt{y}+1)\,dy+\int_{-1}^0 (y-y^3+1) \,dy \\&= \frac{4}{3}\,.
\end{align}
:::
<br>

**Example 12**

Write down an iterated triple integral to calculate the volume of the following region $E$.
![image](https://hackmd.io/_uploads/S1iIyyJ4yx.png)
<br>

:::spoiler Answer

We split the region into two parts: $E_1$ the inverted cone at the bottom and $E_2$ the hemispherical solid at the top. 

Note that at the intersection of $E_1$ and $E_2$ we can substitute  $z=\sqrt{x^2+y^2}$ in $x^2+y^2+z^2=z$, to obtain $z^2+z^2=z$ which gives us $z=1/2$. 

![image](https://hackmd.io/_uploads/HkqJry1E1l.png)

So, the volume can be written as\begin{align}
\text{Volume} (E)
&=\iint_{E_1} 1 \, dx\,dy\,dz + \iint_{E_2} 1 \, dx\,dy\,dz\\ 
&=\int_{-1/2}^{1/2}\int_{-\sqrt{1/4-y^2}}^{\sqrt{1/4-y^2}}\int_{\sqrt{x^2+y^2}}^{1/2} 1\,dz \,dx\,dy\\& \hspace{60pt}+\int_{-1/2}^{1/2}\int_{-\sqrt{1/4-y^2}}^{\sqrt{1/4-y^2}}\int_{1/2}^{1/2+\sqrt{1/4-(x^2+y^2)}} 1\,dz\,dx \,dy
\end{align}

:::
<br>

---
**Reference** 
1. *Chapter 15.1 & 15.6* : Stewart, J. (2012). Calculus (8th ed.). Boston: Cengage Learning. 
2. *Chapter 14.1 & 14.3* : Strang G., Calculus (3rd ed.). [https://ocw.mit.edu/](https://ocw.mit.edu/courses/res-18-001-calculus-fall-2023/pages/textbook/) 
3. *Chapter 3.2 & 3.3* : Corral, M. (2021). Vector Calculus. [https://www.mecmath.net/](https://www.mecmath.net/) 
---
<br>

:::danger

**Summary**: Now, we can

- Define multiple integrals over rectangular and general regions.
- State properties of the integral.
- Change the order of integration for iterated integrals.
- Evaluate multiple integrals over general regions in Cartesian coordinates.
- Use multiple integrals to compute areas and volumes.
- Estimate integrals using bounds on the integrand.
:::