Integrals over trivial regions

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We study integrals over basic regions.

Now we move on to double integrals, and next discuss triple integrals. Instead of integrating over an interval like $[a,b]$ we now integrate over regions like this $R = \{(x,y):\text {$x$ and $y$ satisfy some property}\}$ in particular, in this section we will only considering simple regions, by this we mean rectangles and boxes.

Double integrals

Suppose you have a function $F:\R ^2\to \R$ . A graph of this function is a surface in $\R ^3$ . For example:

We are interested in the “signed volume” between this surface and the $(x,y)$ -plane. This means that the space above the $(x,y)$ plane under $F$ , will have a positive volume. Space above $F$ and under the $(x,y)$ -plane will have a “negative” volume. This is similar to the notion of “signed area” used before. If we want to compute the signed volume of a surface defined by $F$ over a rectangular region, say the rectangle defined by $R = \{(x,y):\text {$a\le x\le b$ and $c\le y\le d$}\}$ we break $R$ into $n$ slices parallel to the $x$ -axis, and $m$ slices parallel to the $y$ -axis. This allows us to consider boxes of dimension $F(x_i^*,y_j^*)\times \underbrace {\left (\frac {b-a}{m}\right )}_{\Delta x}\times \underbrace {\left (\frac {d-c}{n}\right )}_{\Delta y}$ where $(x_i^*,y_j^*)$ is a point in $(i,j)$ -rectangle:

Computing the volume of each of these boxes approximates the signed volume enclosed by the surface:

Letting the number of rectangles in the $x$ -direction and $y$ -direction go to infinity, we will have that $\Delta x \cdot \Delta y$ goes to zero, and we will find the exact volume enclosed by our surface when bounded by the region $R$ . This leads to our definition of a double integral:

Given a function $F:\R ^2\to \R$ , a double integral $\iint _R F(x,y) \d A$ of a function $F$ over a rectangular region $R$ , is given by: $\iint _R F(x,y) \d A = \lim _{\substack {m\to \infty \\ n\to \infty }}\sum _{i=1}^m\sum _{j=1}^n F(x^*_i,y^*_j)\Delta x\Delta y$

Let the value of a function $F:\R ^2\to \R$ be given below:

Let $R = \{(x,y): \text {$0\le x\le 4$ and $0\le y\le 6$}\}$ $\begin{align*} \Delta x &= \answer{1}\\ \Delta y &= \answer{2} \end{align*}$ Compute $\iint _R F(x,y)\d A$ . $\iint _R F(x,y) \d A = \answer {28}$

How do we compute a double integral with calculus? We use an iterated integral.

Fubini’s Theorem Let $F$ be continuous on the region $R = \{(x,y):\text {$a\le x\le b$ and $c\le y\le d$}\}.$ Then: $\begin{align*} \iint_R F(x,y) \d A &= \int_a^b\int_c^d F(x,y)\d y\d x\\ &=\int_c^d\int_a^b F(x,y)\d x\d y. \end{align*}$

This theorem says two things at once:

Double integrals can be computed via iterated integrals.
The order of integration can be changed, as long as the bounds are changed as well.

Now let’s work some examples:

Let $F(x,y) = xy+e^y$ . Find the signed volume under $F$ on the region $R = \{(x,y):\text {$3\le x\le 4$ and $1\le y\le 2$}\}.$

We will compute this integral two different ways. Write with me, $\begin{align*} \iint_R \big(xy+e^y\big) \d A &= \int_3^{\answer[given]{4}}\int_{\answer[given]{1}}^{\answer[given]{2}}\big(xy+e^y\big) \d \answer[given]{y} \d \answer[given]{x}\\ &= \int_{\answer[given]{3}}^{\answer[given]{4}} \eval{\answer[given]{\frac{1}{2}xy^2+e^y}}_{\answer[given]{1}}^{\answer[given]{2}} \d \answer[given]{x} \\ &= \int_{\answer[given]{3}}^{\answer[given]{4}}\left(\answer[given]{\frac{3}{2}x + e^2-e}\right)\d \answer[given]{x} \\ &= \eval{\answer[given]{\frac{3}{4}x^2 + (e^2-e)x}}_{\answer[given]{3}}^{\answer[given]{4}} \\ &= \answer[given]{\frac{21}{4}+ e^2-e}. \end{align*}$ Now let’s compute this integral using a different order of integration. Write with me, $\begin{align*} \iint_R\big(xy+e^y\big) \d A &= \int_1^{\answer[given]{2}}\int_{\answer[given]{3}}^{\answer[given]{4}}\big(xy+e^y\big)\d \answer[given]{x} \d \answer[given]{y} \\ &= \int_{\answer[given]{1}}^{\answer[given]{2}}\eval{\answer[given]{\frac{1}{2}x^2y+xe^y}}_{\answer[given]{3}}^{\answer[given]{4}}\d \answer[given]{y}\\ &= \int_{\answer[given]{1}}^{\answer[given]{2}}\left(\answer[given]{\frac{7}{2}y+e^y}\right)\d \answer[given]{y}\\ &= \eval{\answer[given]{\frac{7}{4}y^2+e^y}}_{\answer[given]{1}}^{\answer[given]{2}}\\ &=\answer[given]{\frac{21}{4}+e^2-e}. \end{align*}$

In our next example, we will see that it is sometimes easier to apply Fubini’s Theorem and integrate with respect to one variable or the other.

Let $F(x,y) = xe^{xy}$ . Find the signed volume under $F$ on the region $R = \{(x,y):\text {$0\le x\le 1$ and $0\le y\le 1$}\}.$

Let’s first compute: $\int _0^1 \int _0^1 x e^{xy} \d x \d y$ As we will see, the difficulty is three-fold. To integrate with respect to $x$ , you can use our first trick, integration by parts: $\begin{align*} \int_0^1 \int_0^1 x e^{xy} \d x \d y &= \int_0^1 \eval{\answer[given]{\frac{xe^{xy}}{y}-\frac{e^{xy}}{y^2}}}_0^1\d y\\ &= \int_0^1 \left(\frac{e^{y}}{y}-\frac{e^y}{y^2} + \frac{1}{y^2}\right)\d y \end{align*}$ Now we notice that this is an improper integral! We must therefore compute $\lim _{b\to \answer [given]{0}}\int _{\answer [given]{b}}^{\answer [given]{1}} \left (\answer [given]{\frac {e^{y}}{y}-\frac {e^y}{y^2} + \frac {1}{y^2}}\right )\d y$ Now we must integrate with respect to $y$ . This is also tricky. Our second trick is to rewrite our current integrand as $\frac {y e^y - \answer [given]{(e^y -1)}}{y^2}$ and now “see” that this results from the quotient rule being applied to $\frac {e^y-\answer [given]{1}}{y}.$ So now we must compute: $\begin{align*} \lim_{b\to \answer[given]{0}}\eval{\frac{e^y-1}{y}}_{\answer[given]{b}}^{\answer[given]{1}} &= \lim_{\answer[given]{b}\to \answer[given]{0}}\left(\answer[given]{\frac{e^1-1}{1}-\frac{e^b-1}{b}}\right)\\ &= \frac{e^1-1}{1}-\lim_{b\to 0}\answer[given]{\frac{e^b-1}{b}}. \end{align*}$ Now, for our third trick, we’ll use L’Hôpital’s rule. Note that the numerator and denominator both go to zero and are differentiable, so this is OK. $\begin{align*} \lim_{b\to 0}\frac{e^b-1}{b} &= \lim_{b\to 0}\answer[given]{\frac{e^b}{1}}\\ &= \answer[given]{1}. \end{align*}$ So putting this all together, we have $\int _0^1 \int _0^1 x e^{xy} \d x \d y = \answer [given]{e-2}$ Whew. This was hard! But there is an easier way. Use Fubini’s Theorem and integrate with respect to $y$ first! Fubini’s theorem says: $\int _0^1 \int _0^1 x e^{xy} \d x \d y = \int _0^1 \int _0^1 x e^{xy} \d y \d x$ provided you switch the order of integration. Note, in this case the limits of integration for both $x$ and $y$ are the same, but we did switch them. Let’s get on with it! Write with me: $\begin{align*} \int_0^1 \int_0^1 x e^{xy} \d y \d x &= \int_0^1 \eval{\answer[given]{e^{xy}}}_0^1\d x\\ &= \int_0^1 \left(\answer[given]{e^x - 1}\right) \d x \\ &= \eval{\answer[given]{e^x - x}}_0^1\\ &= \answer[given]{e-2} \end{align*}$ Man alive! That was easy!

Triple integrals

Using a similar technique to how we made boxes to define double integrals, we can make four-dimensional boxes to define a triple integral that computes the signed hypervolume bounded by a hypersurface and a three-dimensional region. $\iiint _R F(x,y,z) \d V = \lim _{\substack {\l \to \infty \\ m\to \infty \\ n\to \infty }}\sum _{i=1}^\l \sum _{j=1}^m \sum _{k=1}^n F(x^*_i,y^*_j,z_k^*)\Delta x\Delta y\Delta z$ How do we compute a triple integral with calculus? We use Fubini’s theorem again, along with iterated integrals.

Fubini’s Theorem Let $F$ be continuous on the region $R = \{(x,y,z):\text {$a\le x\le b$, $c\le y\le d$, $p\le z\le q$}\}.$ Then: $\iiint _R F(x,y,z) \d V = \int _a^b\int _c^d\int _p^q F(x,y,z)\d z \d y\d x.$ As before, all six combinations of orders of integration will work, provided that the bounds are changed appropriately.

Let $F$ be a continuous function on the region $R$ . $\begin{align*} \int_a^b\int_c^d\int_p^q &F(x,y,z)\d z \d y\d x \\ &= \int_{\answer{p}}^{\answer{q}}\int_a^{\answer{b}} \int_{\answer{c}}^d F(x,y,z)\d \answer{y} \d \answer{x} \d \answer{z} \end{align*}$

$\int _a^b f(x) \d x = - \int _b^a f(x) \d x$

$\begin{align*} \int_a^b\int_c^d\int_p^q &F(x,y,z)\d z \d y\d x \\ &= \int_{\answer{c}}^{d} \int_{\answer{b}}^{a}\int_{\answer{q}}^{\answer{p}} F(x,y,z)\d \answer{z} \d \answer{x} \d \answer{y} \end{align*}$

Let’s do an example.

Let $F(x,y,z) = \cos (x-y)\sin (z)$ . Find the signed hypervolume “under” $F$ on the region $R = \{(x,y):\text {$0\le x\le \pi $, $0\le y\le \pi $, and $0\le z\le \pi $}\}.$

Write with me: $\begin{align*} \iiint_R F(x,y,z) \d V &=\int_{\answer[given]{0}}^{\answer[given]{\pi}}\int_{\answer[given]{0}}^{\answer[given]{\pi}}\int_{\answer[given]{0}}^{\answer[given]{\pi}} \answer[given]{\cos(x-y)\sin(z)} \d x \d y \d z \\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi}}\int_{\answer[given]{0}}^{\answer[given]{\pi}}\eval{\answer[given]{\sin(x-y)\sin(z)}}_{\answer[given]{0}}^{\answer[given]{\pi}}\d y \d z \\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi}}\int_{\answer[given]{0}}^{\answer[given]{\pi}}\left(\answer[given]{\sin(\pi-y)\sin(z) + \sin(y)\sin(z)}\right)\d y \d z \\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi}}\eval{\answer[given]{\cos(\pi-y)\sin(z) -\cos(y)\sin(z)}}_{\answer[given]{0}}^{\answer[given]{\pi}} \d z \\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi}}\left(\answer[given]{4\sin(z)}\right) \d z \\ &= \eval{\answer[given]{-4\cos(z)}}_{\answer[given]{0}}^{\answer[given]{\pi}} \\ &= \answer[given]{8} \end{align*}$

We’ve just begun our journey with integrals. Next, we’ll think about more complex regions! For some interesting extra reading check out: