Part 11 - General regions (2)

Introduction

In this part we’ll cover how to compute volumes using double integrals, as well as defining the triple integral over general regions.

Properties of double integral

$\iint_D (f(x, y) + g(x, y)) \, dA = \iint_D f(x, y) \, dA + \iint_D g(x, y) \, dA$

Quick proof:

\iint_D (f(x, y) + g(x, y)) \, dA = \lim_{\substack{\Delta x \to 0 \newline \Delta y \to 0}} \sum_{j = 1}^{m} \sum_{i = 1}^{n} (f(x_{i}^{*}, y_{j}^{*}) + g(x_{i}^{*}, y_{j}^{*})) (\Delta A)_{ij}

We can split this sum into:

\sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{*}, y_{j}^{*}) (\Delta A)_{ij}

\sum_{j = 1}^{m} \sum_{i = 1}^{n} g(x_{i}^{*}, y_{j}^{*}) (\Delta A)_{ij}

These are the respective double integrals. Note that all of these properties use the same approach for the proof so :].

$\iint_D c \cdot f(x, y) \, dA = c \cdot \iint_D f(x, y) \, dA$
If $f(x, y) \leq g(x, y)$ for all $(x, y) \in D$ . $\iint_D f(x, y) \, dA \leq \iint_D g(x, y) \, dA$
Suppose $D = D_1 \cup D_2$ where $D_1$ and $D_2$ do not overlap, except at boundary points. $\iint_D f(x, y) \, dA = \iint_{D_1} f(x, y) \, dA + \iint_{D_2} f(x, y) \, dA$
$\iint_D k \, dA$ This is the volume under the graph of the constant function, plane, above $D$ . In other words, cylinder of height $k$ with waist $D$ . Which means we have an area of $D \cdot k$ .

In the special case $k = 1$ , we get the area of $D$ .

Computing volume

To understand how we can compute volumes with double integrals, let’s do this example.

Example 1

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 line, $y = 2x$ and parabola $y = x^2$ .

So, $f(x, y) = x^2 + y^2$ . We have a region bounded by functions of $x$ , which means we have a type I region.

Let’s find the intersection points:

\begin{align*} 2x & = x^2 \newline x^2 - 2x & = 0 \newline x(x - 2) & = 0 \newline x_1 & = 0 \newline x_2 & = 2 \end{align*}

We want to compute the volume under the paraboloid. This means:

\iint_D f(x, y) \, dA = \int_0^2 \int_{x^2}^{2x} (x^2 + y^2) \, dy \, dx

\int_0^2 \left(x^2y + \dfrac{y^3}{3}\right) \bigg\rvert_{y = x^2}^{y = 2x} \, dx

\int_0^2 \left[\left(2x^3 + \dfrac{8x^3}{3}\right) - \left(x^4 + \dfrac{x^6}{3}\right)\right] \, dx

\int_0^2 \left(\dfrac{14}{3} x^3 - x^4 - \dfrac{x^6}{3}\right) \, dx

\left(\dfrac{14}{12} x^4 - \dfrac{x^5}{5} - \dfrac{x^7}{21}\right) \bigg\rvert_{x = 0}^{x = 2} = \ldots = \dfrac{56}{3} - \dfrac{32}{5} - \dfrac{128}{21}

Let’s do a more challenging example.

Example 2

Find the volume of the tetrahedron, bounded by the planes, $x + 2y + z = 2, x = 2y, x = 0, z = 0$ .

Let’s decipher this a bit. $x = 0$ , just means $yz$ -plane and $z = 0$ just means $xy$ -plane.

But we need to find the points on the plane $x + 2y + z = 2$

It is sufficient to find three points.

Let’s set $x = 0$ and $y = 0$ (meaning $z$ -axis):

z = 2

Which means point $(0, 0, 2)$ .

Now set $x = 0$ and $z = 0$ (meaning $y$ -axis):

\begin{align*} 2y & = 2 \newline y & = 1 \end{align*}

Which means point $(0, 1, 0)$

Now with $x = 2y$ :

\begin{align*} x + x & = 2 \newline 2x & = 2 \newline x & = 1 \end{align*}

Which means $y = \dfrac{1}{2}$

Which means $\left(1, \dfrac{1}{2}, 0\right)$

Now let’s express this plane in terms of two variables:

f(x, y) = 2 - x - 2y

We call this region for $D$ . Our tetrahedron lies under the graph for $f(x, y)$ above $D$ .

We need to find the line that holds our region. Since we know 2 points on this plane, let’s use these.

y = kx + m

k = \dfrac{\Delta y}{\Delta x} = \dfrac{1 - \dfrac{1}{2}}{0 - 1} = -\dfrac{1}{2}

\begin{align*} 1 & = -\dfrac{1}{2} \cdot 0 + m \newline m & = 1 \end{align*}

y = -\dfrac{1}{2} x + 1

We have a bounded region with functions of $x$ , this means type I region.

Therefore, the volume must be:

\int_0^1 \int_{\frac{x}{2}}^{-\frac{x}{2} + 1} (2 - x - 2y) \, dy \, dx

I’ll save myself some time (for now) and just give us the final answer:

\boxed{\dfrac{1}{3}}

Triple integrals over general regions

We know how to compute the $\iiint$ over a rectangular box. Let’s use the same approach as for general regions for the double integral.

$f(x, y, z)$ is defined on some bounded solid, $E$ . Let’s enclose $E$ in a rectangular box and define $\tilde{f}$ .

Let’s call this rectangular box for $B$ .

\tilde{f} = \begin{cases} f(x, y, z) & (x, y, z) \in E \newline 0 & (x, y, z) \notin E \end{cases}

This means that:

\iiint_E f(x, y, z) \, dV = \iiint_B \tilde{f}(x, y, z) \, dV

The main idea here is that, when computing several integrals, we can view our variables as “inner” and “outer” variables.

In the case for computing double integrals, we only have one inner and one outer, so it’s not that interesting.

However, in the case for three or more variables, we must have this view.

If we for example make $z$ the inner variable and $x$ and $y$ the outer variables then:

\iiint_E f(x, y, z) \, dV = \iint_D \int_{g_1(x, y)}^{g_2(x, y)} f(x, y, z) \, dz \, dA

Let’s properly define it.

Definition 1 (Triple integral over a general region)

To find the triple integral, over solid $E$ . Choose one inner variable and two outer variables. Fix the outer variables, then determine how the inner variable changes. Then integrate, at first over inner variable, then over outer variables.

\iiint_E f(x, y, z) \, dV = \iint_D \int_{g_1(x, y)}^{g_2(x, y)} f(x, y, z) \, dz \, dA

Example 3

Find $\iiint_E x \, dV$ , where $E$ is the ball $x^2 + y^2 + z^2 \leq 1$ . $E$ lies between the upper hemisphere and lower hemisphere above the unit disc, with radius 1 in the $xy$ -plane.

Let’s see how $z$ changes:

\begin{align*} z^2 & = 1 - x^2 - y^2 \newline z & = \pm \sqrt{1 - x^2 - y^2} \end{align*}

E = \{(x, y, z) \mid (x, y) \in D, -\sqrt{1 - x^2 - y^2} \leq z \leq \sqrt{1 - x^2 - y^2} \}

Let’s make $z$ our inner variable:

\iiint_E x \, dV = \iint_D \int_{-\sqrt{1 - x^2 - y^2}}^{\sqrt{1 - x^2 - y^2}} x \, dz \, dA

\iint_D xz \bigg\rvert_{z = -\sqrt{1 - x^2 - y^2}}^{z = \sqrt{1 - x^2 - y^2}} \, dA

\iint_D 2x\sqrt{1 - x^2 - y^2} \, dA

Let’s find this double integral. Looking at this integral, one can see that, if we integrate with respect to $x$ first, it will be a lot easier.

Since the unit disc is of both type I and II, let’s perform a type II double integral.

\int_{-1}^{1} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} 2x\sqrt{1 - x^2 - y^2} \, dx \, dy

Let $u = x^2$ and therefore $du = 2x \, dx$ . Our new limits are therefore $1 - y^2$ .

\int_{-1}^{1} \int_{1 - y^2}^{1 - y^2} \sqrt{1 - y^2 - u} \, du \, dy

We have an integral from and to the same limit, therefore this entire integral becomes 0.

Properties of triple integrals

As we have seen with double integrals, we have some properties.

1) - 4) same as double integral properties.

However: 5) $\iint_D 1 \, dA$ = Area over $D$ .

In the triple integral case: 5) $\iiint_E 1 \, dV$ = Volume over solid.