(Problem 1) Compute where is the rectangle with and .
In this section we define the double integral over a rectangle.
Recall that for a function of one variable defined and continuous on an interval , we have the definite integral In other words, the definite integral is defined as a limit of Riemann Sums. The points are called sample points and they are distributed fairly evenly throughout the interval .
For double integrals, we modify this approach by considering a rectangle in the domain of continuity of a function of two variables, . Let be the rectangle defined by
Double Integral Let be a continuous function defined on the rectangle described
above. The double integral of over is defined by where the points are sample
points roughly evenly distributed in the rectangle .
The double integral is thus defined as the limit of a double Riemann sum.
The double integral of a positive function gives the volume of the solid situated
above the rectangle and beneath the surface .
Example 1 Compute where is the rectangle with and .
The function is continuous and positive on the rectangle . Since the function is constant and is a rectangle, the solid contained above and beneath the graph of is a rectangular solid whose volume is length width height. Hence the value of the double integral is
The function is continuous and positive on the rectangle . Since the function is constant and is a rectangle, the solid contained above and beneath the graph of is a rectangular solid whose volume is length width height. Hence the value of the double integral is
Example 2 Compute where is the square with and .
The function is continuous and positive on the rectangle . To determine the nature of the surface , it is helpful to square both sides and rewrite as This is a circle of radius centered at the origin of the -plane. In this is a surface type called a cylinder (as discussed in section 1.8). In fact, this particular surface is a right circular cylinder of radius whose central axis is the -axis. The surface intersects the -plane in the band which exactly matches the -dimension of the square . Thus the double integral in question is asking for the volume of half (since ) of this cylinder from to . Thus
The function is continuous and positive on the rectangle . To determine the nature of the surface , it is helpful to square both sides and rewrite as This is a circle of radius centered at the origin of the -plane. In this is a surface type called a cylinder (as discussed in section 1.8). In fact, this particular surface is a right circular cylinder of radius whose central axis is the -axis. The surface intersects the -plane in the band which exactly matches the -dimension of the square . Thus the double integral in question is asking for the volume of half (since ) of this cylinder from to . Thus
Here is a video solution of problem 2:
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If the function over the region , then the double integral and its value represents the
negative of the volume of the associated region.
Example 3 Compute where is the rectangle with and .
The function is continuous and negative on the rectangle . Since the function is constant and is a rectangle, the solid contained below and above the graph of is a rectangular solid whose volume is length width height. Hence the value of the double integral is the negative of this volume
The function is continuous and negative on the rectangle . Since the function is constant and is a rectangle, the solid contained below and above the graph of is a rectangular solid whose volume is length width height. Hence the value of the double integral is the negative of this volume
We now consider double integrals that we are not able to compute from purely geometric considerations.
0.1 Iterated Integrals
The second and third integrals in Fubini’s Theorem are called iterated integrals
and they are computed in a manner similar to second order mixed partial derivatives.
We compute the inner integral first, treating the variable of the outer integral as a
constant. Then we compute the outer integral.
Example 4 Compute where is the rectangle with and .
The function is continuous on the rectangle , so we can use Fubini’s Theorem to compute the double integral. We have \begin{align*} \iint _R (x^2 + y^2) \, dA &= \int _0^2 \int _0^3 (x^2 + y^2) \, dy\, dx\\ &= \int _0^2 \left (x^2 y + \frac 13 y^3 \bigg |_0^3 \right ) \, dx\\ &= \int _0^2 \left [(3x^2 + 9) - (0+0) \right ] \, dx\\ &= \int _0^2 (3x^2 + 9) \, dx\\ &= x^3 + 9x \bigg |_0^2\\ &= 26 \end{align*}
The function is continuous on the rectangle , so we can use Fubini’s Theorem to compute the double integral. We have \begin{align*} \iint _R (x^2 + y^2) \, dA &= \int _0^2 \int _0^3 (x^2 + y^2) \, dy\, dx\\ &= \int _0^2 \left (x^2 y + \frac 13 y^3 \bigg |_0^3 \right ) \, dx\\ &= \int _0^2 \left [(3x^2 + 9) - (0+0) \right ] \, dx\\ &= \int _0^2 (3x^2 + 9) \, dx\\ &= x^3 + 9x \bigg |_0^2\\ &= 26 \end{align*}
We can also compute the iterated integral in the other order: \begin{align*} \iint _R (x^2 + y^2) \, dA &= \int _0^3 \int _0^2 (x^2 + y^2) \, dx\, dy\\ &= \int _0^3 \left ( \frac 13 x^3 + xy^2 \bigg |_0^2 \right ) \, dy\\ &= \int _0^3 \left (\frac 83 + 2y^2 \right ) \, dy\\ &= \frac 83 y + \frac 23 y^3 \bigg |_0^3\\ &= 8 + 18 = 26 \end{align*}
Here is a video solution of problem 4:
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Example 5 Compute where is the unit square with and .
The function is continuous on the unit square, so we can use Fubini’s Theorem to compute the double integral. We have \begin{align*} \iint _R (x^3y^2 - 4xy) \, dA &= \int _0^1 \int _0^1 (x^3y^2 - 4xy) \, dy\, dx\\ &= \int _0^1 \left (\frac 13 x^3 y^3 - 2xy^2 \bigg |_0^1 \right ) \, dx\\ &= \int _0^1 \left (\frac 13 x^3 - 2x \right ) \, dx\\ &= \frac{1}{12} x^4 - x^2 \bigg |_0^1\\ &= -\frac{11}{12} \end{align*}
The function is continuous on the unit square, so we can use Fubini’s Theorem to compute the double integral. We have \begin{align*} \iint _R (x^3y^2 - 4xy) \, dA &= \int _0^1 \int _0^1 (x^3y^2 - 4xy) \, dy\, dx\\ &= \int _0^1 \left (\frac 13 x^3 y^3 - 2xy^2 \bigg |_0^1 \right ) \, dx\\ &= \int _0^1 \left (\frac 13 x^3 - 2x \right ) \, dx\\ &= \frac{1}{12} x^4 - x^2 \bigg |_0^1\\ &= -\frac{11}{12} \end{align*}
In general the value of a double integral gives the volume of the region above the
-plane minus the volume of the region below the -plane.
Example 6 Compute where is the unit square with and .
The function is continuous on the unit square, so we can use Fubini’s Theorem to compute the double integral. Thus, the double integral can be written as either and for this integral it makes a difference. Since the integral can be computed using -substitution and the integral requires integration by parts, the first option is to be preferred. With the -substitution , we have \begin{align*} \iint _R xy\cos (x^2y) \, dA &= \int _0^1 \int _0^1 xy\cos (x^2y) \, dx\, dy\\ &= \frac 12 \int _0^1 \int _0^y \cos (u) \, du \, dy\\ &= \frac 12 \int _0^1 \sin (y) \, dy\\ &= -\frac 12 \cos (y) \bigg |_0^1\\ &= \frac 12 \left (1 - \cos (1)\right ) \end{align*}
The function is continuous on the unit square, so we can use Fubini’s Theorem to compute the double integral. Thus, the double integral can be written as either and for this integral it makes a difference. Since the integral can be computed using -substitution and the integral requires integration by parts, the first option is to be preferred. With the -substitution , we have \begin{align*} \iint _R xy\cos (x^2y) \, dA &= \int _0^1 \int _0^1 xy\cos (x^2y) \, dx\, dy\\ &= \frac 12 \int _0^1 \int _0^y \cos (u) \, du \, dy\\ &= \frac 12 \int _0^1 \sin (y) \, dy\\ &= -\frac 12 \cos (y) \bigg |_0^1\\ &= \frac 12 \left (1 - \cos (1)\right ) \end{align*}
Here is a video solution of problem 6:
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