11.3: Double Integrals over General Regions

Example $11.3.3$

Let $f(x,y) = x^2y$ be defined on the triangle $D$ with vertices $(0,0)\text{,}$ $(2,0)\text{,}$ and $(2,3)$ as shown at left in Figure $11.3.3$.

To evaluate $\iint_D f(x,y) \, dA\text{,}$ we must first describe the region $D$ in terms of the variables $x$ and $y\text{.}$ We take two approaches.

Approach 1: Integrate first with respect to $y\text{.}$

In this case we choose to evaluate the double integral as an iterated integral in the form

$$\iint_D x^2y \, dA = \int_{x=a}^{x=b} \int_{y=g_1(x)}^{y=g_2(x)} x^2y \, dy \, dx,$$

and therefore we need to describe $D$ in terms of inequalities

$$g_1(x) \leq y \leq g_2(x) \ \ \ \ \ \text{ and } \ \ \ \ \ a \leq x \leq b.$$

Since we are integrating with respect to $y$ first, the iterated integral has the form

$$\iint_D x^2y \, dA =\int_{x=a}^{x=b} A(x) \, dx,$$

where $A(x)$ is a cross sectional area in the $y$ direction. So we are slicing the domain perpendicular to the $x$-axis and want to understand what a cross sectional area of the overall solid will look like. Several slices of the domain are shown in the middle image in Figure $11.3.4$. On a slice with fixed $x$ value, the $y$ values are bounded below by 0 and above by the $y$ coordinate on the hypotenuse of the right triangle. Thus, $g_1(x) = 0\text{;}$ to find $y = g_2(x)\text{,}$ we need to write the hypotenuse as a function of $x\text{.}$ The hypotenuse connects the points (0,0) and (2,3) and hence has equation $y = \frac{3}{2}x\text{.}$ This gives the upper bound on $y$ as $g_2(x) = \frac{3}{2}x\text{.}$ The leftmost vertical cross section is at $x=0$ and the rightmost one is at $x=2\text{,}$ so we have $a=0$ and $b=2\text{.}$ Therefore,

$$\iint_D x^2y \, dA = \int_{x=0}^{x=2} \int_{y=0}^{y = \frac32 x} x^2y \, dy \, dx.$$

We evaluate the iterated integral by applying the Fundamental Theorem of Calculus first to the inner integral, and then to the outer one, and find that

$$\begin{align*} \int_{x=0}^{x=2} \int_{y=0}^{y=\frac32 x} x^2y \, dy \, dx & = \int_{x=0}^{x=2} \left[x^2 \cdot \frac{y^2}{2}\right]\biggm|_{y=0}^{y=\frac32 x} \, dx\\[4pt] & = \int_{x=0}^{x=2} \frac{9}{8}x^4\, dx\\[4pt] & = \frac{9}{8}\frac{x^5}{5}\biggm|_{x=0}^{x=2}\\[4pt] & = \left(\frac{9}{8}\right) \left(\frac{32}{5}\right)\\[4pt] & = \frac{36}{5}. \end{align*}$$

Approach 2: Integrate first with respect to $x\text{.}$

In this case, we choose to evaluate the double integral as an iterated integral in the form

$$\iint_D x^2y \, dA = \int_{y=c}^{y=d} \int_{x=h_1(y)}^{x=h_2(y)} x^2y \, dx \, dy$$

and thus need to describe $D$ in terms of inequalities

$$h_1(y) \leq x \leq h_2(y) \ \ \ \ \ \text{ and } \ \ \ \ \ c \leq y \leq d.$$

Since we are integrating with respect to $x$ first, the iterated integral has the form

$$\iint_D x^2y \, dA = \int_c^d A(y) \, dy,$$

where $A(y)$ is a cross sectional area of the solid in the $x$ direction. Several slices of the domain — perpendicular to the $y$-axis — are shown at right in Figure $11.3.3$. On a slice with fixed $y$ value, the $x$ values are bounded below by the $x$ coordinate on the hypotenuse of the right triangle and above by 2. So $h_2(y) = 2\text{;}$ to find $h_1(y)\text{,}$ we need to write the hypotenuse as a function of $y\text{.}$ Solving the earlier equation we have for the hypotenuse ($y = \frac32 x$) for $x$ gives us $x = \frac{2}{3}y\text{.}$ This makes $h_1(y) = \frac{2}{3}y\text{.}$ The lowest horizontal cross section is at $y=0$ and the uppermost one is at $y=3\text{,}$ so we have $c=0$ and $d=3\text{.}$ Therefore,

$$\iint_D x^2y \, dA = \int_{y=0}^{y=3} \int_{x=(2/3)y}^{x=2} x^2y \, dx \, dy.$$

We evaluate the resulting iterated integral as before by twice applying the Fundamental Theorem of Calculus, and find that

$$\begin{align*} \int_{y=0}^{y=3} \int_{x=\frac{2}{3}y}^{2} x^2y \, dx \, dy & = \int_{y=0}^{y=3} \left[\frac{x^3}{3}\right]\biggm|_{x=\frac{2}{3}y}^{x=2}y \, dx\\[4pt] & = \int_{y=0}^{y=3} \left[\frac{8}{3}y - \frac{8}{81}y^4 \right] \, dy\\[4pt] & = \left[\frac{8}{3}\frac{y^2}{2} - \frac{8}{81}\frac{y^5}{5}\right]\biggm|_{y=0}^{y=3}\\[4pt] & = \left(\frac{8}{3}\right) \left(\frac{9}{2}\right) - \left(\frac{8}{81}\right) \left(\frac{243}{5}\right)\\[4pt] & = 12 - \frac{24}{5}\\[4pt] & = \frac{36}{5}. \end{align*}$$

We see, of course, that in the situation where $D$ can be described in two different ways, the order in which we choose to set up and evaluate the double integral doesn't matter, and the same value results in either case.