Double Integrals

A double integral is an integral

$\dint_R f(x, y)\,dx\,dy.$

R is a region in $\real^2$ , and is an integrable function.

Under appropriate conditions --- for example, if f is continuous function --- you can compute a double integral as an iterated integral:

$\dint_R f(x, y)\,dx\,dy = \int_a^b \left(\int_{u(x)}^{v(x)} f(x, y)\,dy\right)\,dx \quad\hbox{or}\quad \dint_R f(x, y)\,dx\,dy = \int_c^d \left(\int_{p(y)}^{q(y)} f(x, y)\,dx\right)\,dy.$

In the first case, the region is described by inequalities

$R = \left\{\matrix{a \le x \le b \cr u(x) \le y \le v(x) \cr}\right\}.$

$\hbox{\epsfysize=1.75in \epsffile{double-integrals-1.eps}} \hskip0.5in \hbox{\epsfysize=1.5in \epsffile{double-integrals-2.eps}}$

In the second case, the region is described by inequalities

$R = \left\{\matrix{c \le y \le d \cr p(y) \le x \le q(y) \cr}\right\}.$

Example. Compute

$\dint_R (4 x - 6 y + 3)\,dx\,dy, \quad\hbox{where}\quad R = \left\{\matrix{0 \le x \le 1 \cr -1 \le y \le 1 \cr}\right\}.$

I may compute the double integral as an iterated integral, integrating with respect to one variable at a time while holding the other variable constant.

Since all the limits are numbers, I can integrate first with respect to x and then with respect to y, or the other way around. In this problem, there is no reason to prefer one order to another.

$\int_{-1}^1 \int_0^1 (4 x - 6 y + 3)\,dx\,dy = \int_{-1}^1 \left[2 x^2 - 6 x y + 3 x\right]_0^1\,dy = \int_{-1}^1 (5 - 6 y)\,dy = \left[5 y - 3 y^2\right]_{-1}^1 = 10.$

Notice how the limits of integration are matched with the integration variable from inside out:

$\hbox{\epsfysize=1in \epsffile{double-integrals-3.eps}}$

Thus, you integrate with respect to x first (holding y constant), then with respect to y.

You might try doing this integral in the other order, with respect to y and then x:

$\int_0^1 \int_{-1}^1 (4 x - 6 y + 3)\,dy\,dx.$

You should get the same answer.

Example. Sketch the region of integration for the double integral:

(a) $\displaystyle \int_{-1}^1 \int_0^2 f(x, y)\,dx\,dy$ .

(b) $\displaystyle \int_0^2 \int_0^{x^2} f(x, y)\,dy\,dx$ .

(a)

$\hbox{\epsfysize=1.75in \epsffile{double-integrals-4.eps}}$

(b)

$\hbox{\epsfysize=1.75in \epsffile{double-integrals-5.eps}}\quad\halmos$

Example. Compute

$\dint_R \cos e^x\,dx\,dy, \quad\hbox{where}\quad R = \left\{\matrix{0 \le x \le 1 \cr 0 \le y \le e^x \cr}\right\}.$

Note that since y has a variable limit, I must integrate with respect to y first, then x. When I integrate with respect to y, I hold x constant. Thus, the term " $\cos e^x$ " is constant with respect to y when I do the first integration.

$\int_0^1 \int_0^{e^x} \cos e^x\,dy\,dy = \int_0^1 \left[y \cos e^x\right]_0^{e^x}\,dx = \int_0^1 e^x \cos e^x\,dx = \int_1^e e^x \cos u \cdot \dfrac{du}{e^x} =$

$\left[u = e^x, \quad du = e^x\,dx, \quad dx = \dfrac{du}{e^x}; \quad x = 0, u = 1; x = 1, u = e\right]$

$\int_1^e \cos u\,du = \left[\sin u\right]_1^e = \sin e - \sin 1 = -0.43068 \ldots.\quad\halmos$

Example. Compute

$\dint_R e^{(3 y - y^3)}\,dx\,dy, \quad\hbox{where}\quad R = \left\{\matrix{y^2 \le x \le 1 \cr -1 \le y \le 1 \cr}\right\}.$

$\int_{-1}^1 \int_{y^2}^1 e^{(3 y - y^3)}\,dx\,dy = \int_{-1}^1 \left[x e^{(3 y - y^3)}\right]_{y^2}^1\,dy = \int_{-1}^1 (1 - y^2) e^{(3 y - y^3)}\,dy = \int_?^? (1 - y^2) e^u \cdot \dfrac{du}{3(1 - y^2)} =$

$\left[u = 3 y - y^3, \quad du = 3(1 - y^2)\,dy, \quad dy = \dfrac{du}{3(1 - y^2)}\right]$

$\dfrac{1}{3} \int_?^? e^u\,du = \dfrac{1}{3} \left[e^u\right]_?^? = \dfrac{1}{3} \left[e^{(3 y - y^3)}\right]_{-1}^1 = \dfrac{1}{3} \left(e^2 - e^{-2}\right) = 2.41790 \ldots.\quad\halmos$

The double integral over a region R of the constant function 1 is just the area of R.

$\dint_R 1\,dx\,dy = \hbox{area}(R).$

Example. Evaluate the integral without computing any antiderivatives:

$\int_0^2 \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \,dy\,dx.$

The region is

$\left\{\matrix{0 \le x \le 2 \cr -\sqrt{4 - x^2} \le y \le \sqrt{4 - x^2} \cr}\right\}$

Note that $y = \pm \sqrt{4 - x^2}$ gives . But we're only looking at the part from to .

$\hbox{\epsfysize=2in \epsffile{double-integrals-6.eps}}$

The region is a half circle of radius 2, whose area is

$\dfrac{1}{2} \pi \cdot 2^2 = 2 \pi.$

Hence,

$\int_0^2 \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \,dy\,dx = 2 \pi.\quad\halmos$

Example. Evaluate the integral without computing any antiderivatives:

$\int_0^4 \int_0^{(12 - 3 x)/4} \,dy\,dx.$

The region is

$\left\{\matrix{0 \le x \le 4 \cr 0 \le y \le \dfrac{12 - 3 x}{4} \cr}\right\}$

Now $y = \dfrac{12 - 3 x}{4}$ gives , a line with x-intercept 4 and y-intercept 3.

$\hbox{\epsfysize=2in \epsffile{double-integrals-7.eps}}$

The region is a triangle in the first quadrant, whose area is

$\int_0^4 \int_0^{(12 - 3 x)/4} \,dy\,dx = \dfrac{1}{2} \cdot 4 \cdot 3 = 6.\quad\halmos$

Multiple integrals satisfy the monotonicity condition: "Bigger functions give bigger integrals".

Proposition. Suppose f and g are integrable on a region R and $f (x) \ge g(x)$ for all $x \in R$ . Then

$\int_R f \ge \int_R g.\quad\halmos$

Example. Suppose $f(x, y) \ge 6 x y$ for all . Use this to obtain a lower bound for

$\int_0^1 \int_0^x f(x, y)\,dy\,dx.$

$\int_0^1 \int_0^x f(x, y)\,dy\,dx \ge \int_0^1 \int_0^x 6 x y\,dy\,dx = \int_0^1 \left[3 x y^2\right]_0^x \,dx = \int_0^1 3 x^3\,dx =$

$\left[\dfrac{3}{4} x^4\right]_0^1 = \dfrac{3}{4}.$

Thus,

$\int_0^1 \int_0^x f(x, y)\,dy\,dx \ge \dfrac{3}{4}.\quad\halmos$

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