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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 15.1 Double Integrals over Rectangles

Subsection 15.1.1 Before Class

Figure 134. Pre-Class Video 1
Figure 135. Pre-Class Video 2

Subsubsection 15.1.1.1 Review: The Definite Integral

Question 15.1.1.
Let \(f(x)\) be a function defined on the interval \([a,b]\text{.}\)
  1. Describe what the definite integral \(\ds \int_a^b f(x)\, dx\) means.
  2. Write the process used to define the definite integral.
  3. Geometrically, what are we using to find the integral?
Solution.
  1. We interpret the definite integral to give the net area under the curve on the interval \([a,b]\text{.}\)
  2. The definite integral is built out of an approximation of the area under the curve. The simplest shape to use as an approximator is a rectangle; so, each approximation needs a base length and a height.
    In order to make the computation easier, we want to make the base lengths consisent- so, we’ll evenly divide the interval by the number of rectangles. So, \(\Delta x = \dfrac{b-a}{n}\text{.}\)
    For any input \(x_i\text{,}\) we can express it arbitrarily as \(x_i = a + i\Delta x\text{,}\) where \(i\) is an indexing term. This means that the function height can be given by \(f(x_i)\text{.}\)
    So, any given approximating rectangle has an area given by \(f(x_i)\Delta x\) (respectively the height and base of the rectangle). In order to approximate the entire area under the curve, we need to add all the rectangles. This gives the expression
    \begin{equation*} \text{Area }\approx \sum_{i=1}^n f(x_i)\Delta x \end{equation*}
    In order to move from approximation to exact answer, we need to analyze what happens as the number of rectangles increases without bound; the limit accomplishes this. So we arrive at the definition of the definite integral:
    \begin{equation*} \ds \int_a^b f(x)\, dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x \end{equation*}
  3. We’re using a rectangle because it’s the simplest shape we can use to find area.

Subsubsection 15.1.1.2 The Double Integral

Question 15.1.2.
Now let \(f(x,y)\) be a function defined on the rectangle \([a,b]\times [c,d]\text{.}\) How could we extend the definition of the definite integral to a function of two variables?
Solution.
Since we are operating in three dimensions now, we can take the same idea and instead of using rectangles, we can use rectangular prisms to approximate.
However, since we’re in three dimensions, we’re not approximating area anymore; we’re approximating volume.
Example 15.1.3.
This example will develop the definition of the definite integral for a function \(f(x,y)\text{.}\) Let \(f(x,y)\) be defined on the region \(R\) given by
\begin{equation*} R = [a,b]\times [c,d] = \lrbrace{(x,y)\in\R^2\mid a\leq x\leq b, c\leq y\leq d} \end{equation*}
and let \(S\) be the solid that lies above \(R\) and under the graph of \(f\text{:}\)
\begin{equation*} S = \lrbrace{(x,y,z)\in\R^3\mid 0\leq z\leq f(x,y), (x,y)\in R} \end{equation*}
  1. What kind of object can we use to approximate the volume of \(S\text{?}\) What sort of expression do we need to approximate the volume?
  2. Can we improve the approximation? How?
  3. Is there a way of converting from an approximation to an exact answer? How?
  4. The definition of the double integral is then:
Solution.
  1. We can use rectangular prisms to approximate volume. We need to create a similar expression to this one:
    \begin{equation*} f(x_i)\Delta x \end{equation*}
    Subdivide \(R\) into rectangles; in the \(x-\)direction, make the length \(\Delta x = \dfrac{b-a}{m}\text{,}\) where \(m\) is the number of subintervals in the \(x-\)direction. In the \(y-\)direction, make the length \(\Delta y = \dfrac{d-c}{n}\text{,}\) where \(n\) is the number of subintervals in the \(y-\)direction.
    Let \((x_{ij},y_{ij})\) be the midpoint of the rectangle \(R_{ij}\) (that is, the rectangle \(i\) steps horizontally and \(j\) steps vertically). The value \(f(x_{ij},y_{ij})\) gives the output on the surface \(S\) which is precisely the height of the prism.
    So, the volume of an individual prism is given by the product of length (\(\Delta x\)), width (\(\Delta y\)), and height (\(f(x_{ij},y_{ij})\)). This gives the approximation
    \begin{equation*} \text{prism volume} = f(x_{ij},y_{ij})\Delta x\Delta y \end{equation*}
    To find the volume of the entire region, we need to add up the volume of each rectangle \(R_{ij}\text{.}\) Since there are two indices, we have to have two summations. This gives us the expression
    \begin{equation*} \text{volume under } S \approx \sum_{j=1}^n\sum_{i=1}^m f(x_{ij},y_{ij})\Delta x\Delta y \end{equation*}
  2. We can improve the approximation by using more and more rectangles, and creating more and more subdivisions by increasing \(m\) and \(n\text{.}\)
  3. Yes; take the limit as \(m\to\infty\) and as \(n\to \infty\text{.}\)
  4. \(\displaystyle \int_c^d\int_a^b f(x,y)\, dx\, dy = \lim_{n\to\infty}\lim_{m\to\infty} \sum_{i=1}^m\sum_{j=1}^n f(x_{ij},y_{ij})\Delta x\Delta y\)
Example 15.1.4.
Approximate the volume of the solid that lies above the square \(R = [0,3]\times[0,3]\) and below the surface \(z = x^2 + y^2\text{.}\)
Divide \(R\) into three equal squares and choose the sample point to be the upper right corner of each square \(R_{ij}\text{.}\)
Solution.
First determine what the region will look like. This is shown in the image below.
The partition of \([0,3]\times[0,3]\) with \(n=m=3\text{;}\) this creates 9 squares. The sample points are noted at the corners, and each square is labeled \(R_{ij}\)
We know that the volume is given by the approximation
\begin{equation*} V\approx \sum_{i=1}^3\sum_{j=1}^3 f(x_{ij},y_{ij})\Delta x\Delta y \end{equation*}
which can be rewritten (without sigma notation) as
\begin{equation*} V\approx f(1,1)\Delta x\Delta y + f(1,2)\Delta x\Delta y + f(1,3)\Delta x\Delta y + f(2,1)\Delta x\Delta y + f(2,2)\Delta x\Delta y + f(2,3)\Delta x\Delta y + f(3,1)\Delta x\Delta y + f(3,2)\Delta x\Delta y + f(3,3)\Delta x\Delta y \end{equation*}
since \(\Delta x = \dfrac{3-0}{3} = 1\) and \(\Delta y = \dfrac{3-0}{3} = 1\text{,}\) \(\Delta x\Delta y = (1)(1) = 1\text{.}\) This means the computation simplifies to
\begin{equation*} V\approx f(1,1)(1) + f(1,2)(1) + f(1,3)(1) + f(2,1)(1) + f(2,2)(1) + f(2,3)(1) + f(3,1)(1) + f(3,2)(1) + f(3,3)(1) \end{equation*}
Computing the values at the given points and substituting gives
\begin{equation*} V\approx (1)(1) + (5)(1) + (10)(1) + (5)(1) + (8)(1) + (13)(1) + (10)(1) + (13)(1) + (18)(1) \end{equation*}
so that we have \(V\approx 84\)
Example 15.1.5.
If \(R = \lrbrace{(x,y)\mid -3\leq x\leq 3, -2\leq y\leq 2}\text{,}\) then evaluate the integral \(\ds \iint_R \sqrt{9-x^2}\,dA\) by interpreting it as a volume.
Solution.
Drawing the surface gives half of a cylinder, along the \(y-\)axis. The volume of the cylinder is \(\pi r^2 h\text{;}\) in this case, \(r = 3\) and \(h = 2-(-2) = 4\text{.}\) Since we’re dealing with half of the cylinder, our volume is given by \(\dfrac{1}{2}\pi r^2 h = \dfrac{1}{2}\cdot (3)^2\cdot 4 = 18\)

Subsection 15.1.2 Pre-Class Activities

Subsection 15.1.3 In Class

Subsubsection 15.1.3.1 Midpoint Rule

Definition 15.1.6. Midpoint Rule (Double Integrals).
If \(\bar{x_i}\) is the midpoint of the interval \([x_{i-1},x_i]\) and \(\bar{y_j}\) is the midpoint of the interval \([y_{j-1},y_j]\text{,}\) then we can approximate the double integral of \(f(x,y)\) over the region \(R\) by
\begin{equation*} \iint_R f(x,y)\, dA \approx \sum_{i=1}^m\sum_{j=1}^n f(\bar{x_i},\bar{y_j})\Delta A \end{equation*}
where \(\Delta A = \Delta x\,\Delta y\)
Example 15.1.7.
Approximate \(\ds \iint_R (x-3y^2)\, dA\text{,}\) where \(R = \lrbrace{(x,y)\mid 0\leq x\leq 2,1\leq y\leq 2}\) and \(m=n=2\text{,}\) using midpoints.
Solution.
Since \(m = 2\text{,}\) we take \(\Delta x = \dfrac{2-0}{2} = 1\) and \(\Delta y = \dfrac{2-1}{2} = \dfrac{1}{2}\text{.}\) This means \(x_1 = 1\text{,}\) \(x_2 = 2\text{,}\) \(y_1 = \dfrac{3}{2}\text{,}\) and \(y_2 = 2\text{.}\)
Since we want midpoints, we have \(\bar{x_1} = \dfrac{1}{2}\text{,}\) \(\bar{x_2} = \dfrac{3}{2}\text{,}\) \(y_1 = \dfrac{5}{4}\text{,}\) and \(\bar{y_2} = \dfrac{7}{4}\) as our points.
Then,
\begin{align*} \ds \iint (x-3y^2\, dA \amp\approx \sum_{i=1}^2\sum_{j=1}^2 f(\bar{x_i},\bar{y_j})\, \Delta A \\ \amp = f(\bar{x_1},\bar{y_1})\Delta A + f(\bar{x_2},\bar{y_1})\Delta A + f(\bar{x_1},\bar{y_2})\Delta A + f(\bar{x_2},\bar{y_2})\Delta A \\ \amp = f\lrpar{\dfrac{1}{2},\dfrac{5}{4}}\Delta A + f\lrpar{\dfrac{1}{2},\dfrac{7}{4}}\Delta A + f\lrpar{\dfrac{3}{2},\dfrac{5}{4}}\Delta A + f\lrpar{\dfrac{3}{2},\dfrac{7}{4}}\Delta A\\ \amp = \lrpar{-\dfrac{67}{16}}\lrpar{\dfrac{1}{2}} + \lrpar{-\dfrac{139}{16}}\lrpar{\dfrac{1}{2}} + \lrpar{-\dfrac{51}{16}}\lrpar{\dfrac{1}{2}} + \lrpar{-\dfrac{123}{16}}\lrpar{\dfrac{1}{2}} \\ \amp = -\dfrac{95}{8} \end{align*}
So our conclusion is that
\begin{equation*} \iint_R (x-3y^2)\, dA \approx -\dfrac{95}{8} = -11.875 \end{equation*}

Subsubsection 15.1.3.2 Iterated Integrals

Now let’s see how to evaluate double integrals.
Example 15.1.8.
Consider the function \(f(x,y)\text{,}\) defined and integrable on the rectangle \(R = [a,b]\times[c,d]\text{.}\)
  1. How can we make sense of the integral \(\ds\int_c^d f(x,y)\, dy\text{?}\) Is it a necessarily a number?
  2. If we set \(A(x) = \ds\int_c^d f(x,y)\, dy\text{,}\) then what does \(\ds \int_a^b A(x)\, dx\) do?
  3. Put it all together. Interpret the expressions
    \begin{equation*} \ds \int_a^b\int_c^d f(x,y)\, dx\,dy \quad \ds \int_c^d\int_a^b f(x,y)\, dy\,dx \end{equation*}
Solution.
  1. Since we’re integrating with respect to \(y\text{,}\) any and all \(x\) terms are considered constant; so while the \(x\) is treated as a number for the integration, the answer isn’t necessarily a number- it could be a function of \(x\text{.}\)
  2. This computes the integral of a single-variable function- something like what we’re used to. Now, this integral returns a strictly numerical value.
  3. In the first integral, we’re integrating first with respect to \(x\) in order to get a function of \(y\text{,}\) then integrating with respect to \(y\) in order to get a numerical value.
    In the second integral, we’re integrating first with respect to \(y\) in order to get a function of \(x\text{,}\) then integrating with respect to \(x\) in order to get an numerical value.
Example 15.1.9.
Let \(f(x,y) = x^2y^3\text{,}\) defined on the rectangle \(R = \lrbrace{(x,y)\mid 0\leq x\leq 3,1\leq y\leq 2}\text{.}\) Evaluate the following:
  1. \(\displaystyle \ds \int_0^3 \int_1^2 f(x,y)\,dy\,dx\)
  2. \(\displaystyle \ds \int_1^2\int_0^3 f(x,y)\,dx\,dy\)
Solution.
  1. \(\displaystyle \ds \int_0^3 \int_1^2 f(x,y)\,dy\,dx = \dfrac{135}{4}\)
  2. \(\displaystyle \ds \int_1^2\int_0^3 f(x,y)\,dx\,dy = \dfrac{135}{4}\)
Example 15.1.10.
Let \(g(x,y) = xe^y\text{,}\) defined on the rectangle \([0,1]\times[-1,1]\text{.}\) Evaluate the following:
  1. \(\displaystyle \ds \iint_R g(x,y)\,dx\,dy\)
  2. \(\displaystyle \ds \iint_R g(x,y)\,dy\,dx\)
Solution.
  1. \(\displaystyle \ds \iint_R g(x,y)\,dx\,dy = \dfrac{e-e^{-1}}{2}\)
  2. \(\displaystyle \ds \iint_R g(x,y)\,dy\,dx = \dfrac{e-e^{-1}}{2}\)
Question 15.1.11.
In the previous two examples, you should see some similarities between both parts. What are those similarities?
Solution.
The answer is same when the order of integration is changed
Example 15.1.13.
Evaluate the double integral \(\ds \iint_R (y-3xy)\, dA\text{,}\) where \(R = \lrbrace{(x,y)\mid -1\leq x\leq 5,0\leq y\leq 1}\)
Solution.
\(\ds \iint_R (y-3xy)\, dA = -15\)
Example 15.1.14.
Let \(f(x,y) =y\cos (xy)\)
  1. Evaluate the double integral \(\ds \int_0^{\pi/4}\int_0^{\pi/2} f(x,y)\, dy\,dx\text{.}\) What do you notice?
  2. Evaluate the double integral \(\ds \int_0^{\pi/2}\int_0^{\pi/4} f(x,y)\, dx\, dy\text{.}\) What do you notice?
Solution.
  1. \(\ds \int_0^{\pi/4}\int_0^{\pi/2} f(x,y)\, dy\,dx = -\dfrac{4}{\pi}\cos\lrpar{\dfrac{\pi^2}{8}}\text{.}\) This is a very difficult integral, requiring multiple uses of integration by parts.
  2. \(\ds \int_0^{\pi/2}\int_0^{\pi/4} f(x,y)\, dx\, dy = -\dfrac{4}{\pi}\cos\lrpar{\dfrac{\pi^2}{8}}\text{.}\) This order was much simpler than the previous one.
Example 15.1.15.
Find the volume of the solid bounded by the elliptic paraboloid \(x^2 + 4y^2 + z = 16\text{,}\) the planes \(x = 2\) and \(y = 2\text{,}\) and the three coordinate planes.
Solution.
\(\ds \int_0^2\int_0^2 (16-x^2-4y^2)\, dx\, dy = 36\)

Subsection 15.1.4 After Class Activities

Example 15.1.16.

Evaluate \(\ds \iint_R \sqrt{5}\, dA\text{,}\) where \(R = \lrbrace{(x,y)\mid 2\leq x\leq 5,-1\leq y\leq 2}\)
Solution.

Example 15.1.17.

Evaluate \(\ds \iint_R (13-6x)\, dA\text{,}\) where \(R = \lrbrace{(x,y)\mid -1\leq x\leq 1, -2\leq y\leq 2}\)
Solution.

Example 15.1.18.

Evaluate \(\ds \int_1^4\int_0^2 (3xy^2 -3y)\, dy\, dx\)
Solution.

Example 15.1.19.

Evaluate \(\ds \int_0^2\int_0^1 (2x+2y)^2\, dx\, dy\)
Solution.
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