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” or “net 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 $$\underbrace {F(x_i^*,y_j^*)}_{\text {height}}\times \underbrace {\left (\frac {b-a}{m}\right )}_{\Delta x}\times \underbrace {\left (\frac {d-c}{n}\right )}_{\Delta y}$$ where $(x_i^*,y_j^*)$ is a point in $(i,j)$-rectangle:

Computing the volume of each of these boxes approximates and summing them together $$\sum _{j=1}^n\sum _{i=1}^m F(x^*_i,y^*_j)\Delta x\Delta y$$ 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:

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

Let $$R = \{(x,y): \text {$0\le x\le 4$ and $0\le y\le 6$}\}$$ $$\begin{align*} \Delta x &= \answer{1}\\ \Delta y &= \answer{2} \end{align*}$$ Now, simply add up the $z$-values found in the rectangles that are within our region and multiply by $\answer {2}$. 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. At this point we will introduce something called Fubini’s Theorem.

Fubini’s Theorem

Fubini’s Theorem gives us a recipe for computing double integrals. In this class, we are going to have many different versions of Fubini’s Theorem. The common factor between all of these theorems is that with each, the “punch-line” will be: $$\underbrace {\iint _R F(x,y) \d A}_{\text {a double integral}} = \underbrace {\int _?^? \int _?^? F(x,y) \d A}_{\text {an iterated integral}}$$ where an iterated integral is nothing more than two applications of our familiar friend/foe: the single integral.

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. First apply Fubini’s Theorem: $$\iint _R \big (xy+e^y\big ) \d A = \int _3^{\answer [given]{4}}\int _{\answer [given]{1}}^{\answer [given]{2}}\big (xy+e^y\big ) \d \answer [given]{y} \d \answer [given]{x}$$ Now integrate with respect to $y$, treating $x$ as a constant: $$\begin{align*} = \int_{\answer[given]{3}}^{\answer[given]{4}} \eval{\answer[given]{\frac{1}{2}xy^2+e^y}}_{\answer[given]{1}}^{\answer[given]{2}} \d \answer[given]{x} \\ &= \int_{\answer[given]{3}}^{\answer[given]{4}}\left(\answer[given]{\frac{3}{2}x + e^2-e}\right)\d \answer[given]{x} \end{align*}$$ Now integrate with respect to $x$, treating $y$ as a constant: $$\begin{align*} &= \eval{\answer[given]{\frac{3}{4}x^2 + (e^2-e)x}}_{\answer[given]{3}}^{\answer[given]{4}} \\ &= \answer[given]{\frac{21}{4}+ e^2-e}. \end{align*}$$

Now let’s compute this integral using a different order of integration. Write with me,

$$\begin{align*} \iint_R\big(xy+e^y\big) \d A &= \int_1^{\answer[given]{2}}\int_{\answer[given]{3}}^{\answer[given]{4}}\big(xy+e^y\big)\d \answer[given]{x} \d \answer[given]{y} \\ &= \int_{\answer[given]{1}}^{\answer[given]{2}}\eval{\answer[given]{\frac{1}{2}x^2y+xe^y}}_{\answer[given]{3}}^{\answer[given]{4}}\d \answer[given]{y}\\ &= \int_{\answer[given]{1}}^{\answer[given]{2}}\left(\answer[given]{\frac{7}{2}y+e^y}\right)\d \answer[given]{y}\\ &= \eval{\answer[given]{\frac{7}{4}y^2+e^y}}_{\answer[given]{1}}^{\answer[given]{2}}\\ &=\answer[given]{\frac{21}{4}+e^2-e}. \end{align*}$$ Note our answers are the same regardless of the order of integration.

In our next example, we will see that it is sometimes easier to apply Fubini’s Theorem and integrate with respect to one variable instead of 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, this integral will require three tricks. To integrate with respect to $x$, you can use 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 $$\frac {y e^y - \left (\answer [given]{e^y -1}\right )}{y^2}$$ 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!

Now 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! Note, we get the same final answer regardless of the order of integration. Thank-you Fubini!

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. $$\iiint _B F(x,y,z) \d V = \lim _{\substack {\l \to \infty \\ m\to \infty \\ n\to \infty }} \sum _{k=1}^n \sum _{j=1}^m \sum _{i=1}^\l F(x^*_i,y^*_j,z_k^*)\Delta x\Delta y\Delta z$$ How do we compute a triple integral with calculus? We use our second version of Fubini’s Theorem. This time, it will say something like:

$$\underbrace {\iiint _B F(x,y,z) \d V}_{\text {a triple integral}} = \underbrace {\int _?^? \int _?^? \int _?^? F(x,y,z) \d V}_{\text {an iterated integral}}$$

Let’s do an example.

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

• Three Aspects of Fubini’s Theorem, J. C. Fisher and J. Shilleto, Mathematics Magazine, February 1986.
• Visualizing Leibniz’s Rule, M. Frantz, Mathematics Magazine, April 2001.