Objectives
- To come
Section 15.2 Double Integrals over General Regions
Subsection 15.2.1 Before Class
Subsubsection 15.2.1.1 General Regions
In general, the regions we deal with aren’t perfect rectangles- so, integration over any plane region needs to be cast in a different light.
Definition 15.2.1. Type I Region.
We say that a plane region \(D\) is type I if it lies between the graphs of two continuous functions of \(x\text{,}\) i.e.
\begin{equation*}
D = \lrbrace{(x,y)\mid a\leq x\leq b, g_1(x)\leq y\leq g_2(x)}
\end{equation*}
where \(g_1\) and \(g_2\) are continuous on \([a,b]\text{.}\)
Definition 15.2.2. Type II Region.
We say that a plane region \(D\) is type II if it lies between the graphs of two continuous functions of \(y\text{,}\) i.e.
\begin{equation*}
D = \lrbrace{(x,y)\mid c\leq y\leq d, h_1(y)\leq x\leq h_2(y)}
\end{equation*}
where \(h_1\) and \(h_2\) are continuous on \([c,d]\text{.}\)
Example 15.2.3.
This example will develop a usable formula for Type I and Type II integrals. Let the region \(R\) be defined by \(R = \lrbrace{(x,y)\mid 1\leq x \leq 2, 2\leq y \leq 4}\text{.}\)
- Sketch the region \(R\text{.}\)
- Consider some function \(f(x,y)\) which is defined on \(R\text{.}\) In the expression \(\ds \int_1^2\int_3^4 f(x,y)\, dy\, dx\text{,}\) what are the bounds communicating to you, in terms of \(R\text{?}\) Rewrite the integral expression and \(R\) to reflect your answer.
- Now sketch what would happen to \(R\) if we replace 3 with the function \(g_1(x) = x^2\text{.}\) What must change in the integral expression? Call the new region \(R_1\text{.}\)
- If we replace 4 with the function \(g_2(x) = -x^2+3x+2\text{,}\) what changes about \(R_1\text{?}\) What about the integral?
- In (c), what would have happened to \(R\) if we replaced 1 with \(h_1(y) = 4-y\) instead? Draw the new region, and call it \(R_2\text{.}\) What change would you see on the integral from part (b)?
- Now what would happen to \(R_2\) if we replace 2 with \(h_2(y) = y^2-4y+4\text{?}\) Sketch the new \(R_2\text{.}\) What change would you see on the integral in part (e)?
Solution.
-
The bounds are communicating functions which define the region.Specifically, we can think of the bounds 1 and 2 as the functions \(a_1(y) = x=1\) and \(a_2(y) = x=2\) (which are functions of \(y\text{,}\) so integrating with respect to \(y\) makes sense). We can also think of 3 and 4 as the functions \(b_1(x) = y=3\) and \(b_2(x) = y = 4\) (which are functions of \(x\text{,}\) so integrating with respect to \(x\) makes sense).This means we can rewrite the integral like this:\begin{equation*} \int_1^2\int_3^4 f(x,y)\, dy\, dx = \int_{x=1}^{x=2}\int_{y=3}^{y=4}f(x,y)\, dy\, dx \end{equation*}
-
Since the bound (function) \(y = 3\) is getting replaced by the bound (function) \(g_1(x) = x^2\text{,}\) the integral must change to reflect the new lower bound:\begin{equation*} \int_{x=1}^{x=2} \int_{g_1(x) = x^2}^{y = 4} f(x,y)\, dx\, dy = \int_1^2\int_{x^2}^4 f(x,y)\, dx\, dy \end{equation*} -
Since the bound (function) \(y = 4\) is getting replaced by the bound (function) \(g_2(x) = -x^2+3x+2\text{,}\) the integral must change to reflect the new upper bound:\begin{equation*} \int_{x=1}^{x=2} \int_{g_1(x) = x^2}^{g_2(x) = -x^2+3x+2} f(x,y)\, dx\, dy = \int_1^2\int_{x^2}^{-x^2+3x+2} f(x,y)\, dx\, dy \end{equation*} -
Since the bound (function) \(x=1\) is getting replaced by the bound (function) \(h_1(y) = 4-y\text{,}\) the integral must change to reflect the new lower bound:\begin{equation*} \int_{h_1(y) = 4-y}^{x=2}\int_{y=3}^{y=4} f(x,y)\, dx \, dy = \int_{4-y}^{2}\int_3^4 f(x,y)\, dx\, dy \end{equation*} -
Since the bound (function) \(x=2\) is getting replaced by the bound (function) \(h_2(y) = y^2-4y+4\text{,}\) the integral must change to reflect the new upper bound:\begin{equation*} \int_{h_1(y) = y+1}^{h_2(y) = y^2-4y+4}\int_{y=1}^{y=2} f(x,y)\, dx \, dy = \int_{4-y}^{y^2-4y+4}\int_3^4 f(x,y)\, dx\, dy \end{equation*}
Integrals of Type I and Type II Regions.
If \(f\) is continuous on a Type I region \(D\) described by
\begin{equation*}
D = \lrbrace{((x,y) \mid a\leq x\leq b, g_1(x)\leq y \leq g_2(x)}
\end{equation*}
then
\begin{equation*}
\iint_D f(x,y)\, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx
\end{equation*}
If \(f\) is continuous on a Type II region \(E\) described by
\begin{equation*}
E = \lrbrace{(x,y)\mid h_1(y)\leq x\leq h_2(y), c\leq y\leq d}
\end{equation*}
then
\begin{equation*}
\iint_E f(x,y)\, dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy
\end{equation*}
Example 15.2.4.
Let \(D\) be the region bounded by the curves \(y = 3x^2\) and \(y =1+2x^2\)
- Sketch and label the curves, and indicate where the region \(D\) is.
- Write the region using the definition of a Type I region.
- Now compute the integral \(\ds \iint_D (x+3y)\, dA\text{.}\)
Solution.
![A sketch of the region \(D\) given in the problem, on the interval \([-2,2]\text{.}\)](generated/latex-image/image-200.svg)
- Formally, we can write the region using set-builder notation as\begin{equation*} D = \lrbrace{(x,y)\mid -1\leq x\leq 1, 3x^2\leq y \leq 1+2x^2} \end{equation*}
-
From part b, we know that\begin{equation*} D = \lrbrace{(x,y)\mid -1\leq x\leq 1, 3x^2\leq y \leq 1+2x^2} \end{equation*}This means we can write the integral as\begin{equation*} \int_{-1}^1 \int_{3x^2}^{1+2x^2} (x+3y)\, dy\, dx \end{equation*}Evaluate from inside out. This means we’re first computing\begin{equation*} \int_{3x^2}^{1+2x^2} (x+3y)\, dy \end{equation*}So we have\begin{align*} \int_{3x^2}^{1+2x^2} (x+3y)\, dy \amp = \lrpar{xy + \dfrac{3}{2}y^2}\bigg\rvert_{y=3x^2}^{y=1+2x^2} \\ \amp = \lrpar{x(1+2x^2) + \dfrac{3}{2}(1+2x^2)} - \lrpar{x(3x^2) + \dfrac{3}{2}(3x^2)} \\ \amp = -\dfrac{15}{2}x^4 -x^3 + 6x^2 + x + \dfrac{3}{2} \end{align*}Since the first integral evaluates as \(-\dfrac{15}{2}x^4 -x^3 + 6x^2 + x + \dfrac{3}{2}\text{,}\) we can rewrite the integral as\begin{equation*} \int_{-1}^1 \int_{3x^2}^{1+2x^2} (x+3y)\, dy\, dx = \int_{-1}^1 \lrpar{-\dfrac{15}{2}x^4 -x^3 + 6x^2 + x + \dfrac{3}{2}}\, dx \end{equation*}which evaluates to 45.
Example 15.2.5.
Find the volume of the solid that lies under the paraboloid \(z = x^2 +2y^2\text{,}\) and above the region \(R\) in the \(xy-\)plane bounded by the curves \(y = x^3\) and \(y = x\text{.}\) Treat \(R\) as a Type I region.
Solution.
Here is a sketch of \(R\text{:}\)
Since we’re treating \(R\) as a Type I region, we need to find a description of the form
\begin{equation*}
R = \lrbrace{(x,y)\mid a\leq x\leq b, g_1(x)\leq y\leq g_2(x)}
\end{equation*}
Based on the region, we can say that \(g_1(x) = x^3\) and \(g_2(x) = x\text{.}\) The \(x-\)values must be \(x=0\) and \(x = 1\) since the curves intersect at those locations. This means our integral is set up as
\begin{equation*}
\int_0^1 \int_{x^3}^x (x^2 + 2y^2)\, dy\, dx
\end{equation*}
So we compute:
\begin{align*}
\int_0^1 \int_{x^3}^x (x^2 + 2y^2)\, dy\, dx \amp = \int_0^1 \lrpar{x^2y + \dfrac{2}{3}y^3}\bigg\rvert_{x^3}^x\\
\amp = \int_0^1 \lrpar{-\dfrac{2}{3}x^9 -x^5 +\dfrac{5}{3}x^3}\, dx\\
\amp = \dfrac{11}{60}
\end{align*}
Example 15.2.6.
Again let \(R\) to be the region in the \(xy-\)plane bounded by the curves \(y = x^3\) and \(y = x\text{.}\)
- Write \(R\) as a Type II region.
- Compute \(\ds \iint_R (x^2 + 2y^2)\, dA\text{.}\) How does your work and answer compare to the previous example?
Solution.
-
For convenience, here is the sketch of \(R\) again:
As a Type II region, we need to express \(R\) in the form\begin{equation*} R = \lrbrace{(x,y)\mid c\leq y\leq d, h_1(y)\leq y\leq h_2(y)} \end{equation*}Rewriting as functions of \(y\text{,}\) \(y = x^3\) becomes \(x = \sqrt[3]{y}\) and \(y = x\) stays \(x = y\text{.}\) Our upper function will be \(h_2(y) = \sqrt[3]{y}\) and our lower function will be \(h_1(y) = y\text{.}\)For our \(y-\)coordinate bounds, we still look at the intersection. There, the \(y-\)coordinates are \(y = 0\) and \(y = 1\text{.}\) -
\begin{align*} \iint_R (x^2 + 2y^2)\, dA \amp = \int_0^1 \int_{y}^{y^{1/3}} (x^2 + 2y^2)\, dx\, dy\\ \amp = \int_0^1 \lrpar{\dfrac{1}{3}y + \dfrac{2}{3}y^{7/3} - y^3}\, dy\\ \amp = \lrpar{\dfrac{1}{6}y^2 + \dfrac{1}{5}y^{10/3} - \dfrac{1}{4}y^4}\bigg\rvert_0^1\\ \amp = \dfrac{11}{60} \end{align*}The work is similar but distinct. Since our bounds changed, the integral itself looks substantially different after the first integral. However, the answer is the same- which we would expect (and hope) since the region and function are the same.
Subsubsection 15.2.1.2 Properties of Double Integrals
Double integrals, because they are integrals, share most of the properties of single integrals. We’ll isolate a few which can be helpful (without proof):
- \(\displaystyle \ds \iint_D [f(x,y) \pm g(x,y)]\, dA = \iint f(x,y)\, dA \pm \iint_D g(x,y)\, dA\)
- \(\displaystyle \ds \iint_D c\cdot f(x,y)\, dA = c\cdot \iint_D f(x,y)\, dA\)
- If \(f(x,y) \leq g(x,y)\) for all \((x,y)\) in \(D\text{,}\) then\begin{equation*} \iint_D f(x,y)\, dA \leq \iint_D g(x,y)\, dA \end{equation*}
- Let \(D_1\) and \(D_2\) be two non-overlapping regions (except potentially on their common boundary). Then, \(D = D_1\cup D_2\) and\begin{equation*} \iint_D f(x,y)\, dA = \iint_{D_1} f(x,y)\, dA + \iint_{D_2} f(x,y)\, dA \end{equation*}
- \(\displaystyle \ds \iint_D 1\, dA = \text{Area}(D)\)
Subsection 15.2.2 Pre-Class Activities
Subsection 15.2.3 In Class
Subsubsection 15.2.3.1 Some Examples
Example 15.2.7.
Evaluate \(\ds \iint_D x^2y^2\, dA\text{,}\) where \(D\) is the region bounded by the curves \(y = x\) and \(y^2 = x+6\text{.}\)
Solution.
The region \(D\) is shown below:
If we try to treat this as a Type I region, we’ll need to split into two integrals (corresponding to the two different lower bounds). So, instead treat \(D\) as a Type II region:
\begin{equation*}
D = \lrbrace{(x,y)\mid -2\leq y\leq 3, y^2-6\leq x \leq y}
\end{equation*}
This gives the integral
\begin{equation*}
\int_{-2}^3\int_{y^2-6}^{y} x^2y^2\, dx\, dy
\end{equation*}
which evaluates to \(\dfrac{50375}{378}\)
Example 15.2.8.
Find the volume of the first-octant tetrahedron bounded by the plane \(x + 2y + z = 2\text{.}\)
Solution.
To determine the region of integration, sketch the problem in three-dimensional space (I don’t know how to do this right now, so I’ll have to come back and add the image later). Doing this gives \(f(x,y) = 2-x-2y\) and \(R = \lrbrace{(x,y)\mid 0\leq x\leq 2, 0\leq y\leq -\dfrac{1}{2}x+1}\) (as a Type I region) or \(R = \lrbrace{(x,y)\mid 0\leq y\leq 1, 0\leq x\leq 2-2y}\) (as a Type II region).
The Type I integral setup is given by
\begin{equation*}
\int_0^2\int_0^{-x/2 + 1} (2-x-2y)\, dy\, dx
\end{equation*}
and the Type II integral setup is given by
\begin{equation*}
\int_0^1\int_0^{2-2y} (2-x-2y)\, dx\, dy
\end{equation*}
In either event, the integral has a value of \(\dfrac{2}{3}\)
Example 15.2.9.
Find the volume of the solid created by the function \(f(x,y) = \dfrac{y}{x^2+1}\) over the region \(R\) bounded by the curves \(y = \sqrt{x}\text{,}\) \(y = 0\text{,}\) and \(x = 4\text{.}\)
Solution.
The Type I integral looks like
\begin{equation*}
\int_0^4\int_0^{\sqrt{x}} \dfrac{y}{x^2+1}\, dy\, dx
\end{equation*}
and the Type II integral takes the form
\begin{equation*}
\int_0^2\int_{y^2}^{4} \dfrac{y}{x^2+1}\, dx\, dy
\end{equation*}
and the value is \(\dfrac{1}{4}\ln 17\)
Example 15.2.10.
Consider the integral \(\ds \int_0^2 \int_0^{y^2} x^2y\, dx\, dy\)
- Sketch and label the region described in the integral.
- Evaluate the integral.
Solution.
Consider the integral \(\ds \int_0^2 \int_0^{y^2} x^2y\, dx\, dy\)
- As a Type I integral, we have\begin{equation*} \int_0^4 \int_{\sqrt{x}}^2 (x^2y)\, dy\, dx \end{equation*}and as a Type II integral we have\begin{equation*} \int_0^2 \int_0^{y^2}(x^2y)\, dx\, dy \end{equation*}which evaluates to \(\dfrac{32}{3}\)
Subsubsection 15.2.3.2 Changing the Order of Integration
Example 15.2.11.
Let \(f(x,y) = \sec^2(x^2)\)
- Compute \(\ds \int_0^1\int_y^1 f(x,y)\, dx\, dy\text{.}\) You should run into some issues; what are they?
- Draw and label the region described in part (a).
- Use part (b) to rewrite the integral in part (a), then evaluate \(\ds \iint_D f(x,y)\, dA\)
Solution.
- The problem is immediately that we don’t have a technique to integrate \(\sec^2(x^2)\text{.}\)
- As written, the integral is Type II; as a Type I integral, the region is described by \(R = \lrbrace{(x,y)\mid 0\leq x\leq 1, 0\leq y\leq x}\text{.}\) This means we can rewrite the integral as\begin{equation*} \int_0^1 \int_0^x \sec^2(x^2)\, dy\, dx \end{equation*}Evaluating the first integral now gives\begin{equation*} \int_0^1 x\sec^2(x^2)\, dx \end{equation*}which can be evaluated using substitution. This gives a value of \(\dfrac{1}{2}\tan(1)\text{.}\)
Example 15.2.12.
Evaluate \(\ds \iint_D e^{-y^2}\, dA\text{,}\) where \(D = \lrbrace{(x,y)\mid 0\leq x\leq 3, x\leq y \leq 3}\)
Solution.
The Type II region is not feasible because there is no antiderivative for \(e^{-y^2}\text{.}\) But after drawing and rewriting as a Type I region, we have
\begin{equation*}
\end{equation*}
So the integral becomes
\begin{equation*}
\int_0^3\int_0^y e^{-y^2}\, dx\, dy
\end{equation*}
whose value is \(\dfrac{1-e^{-9}}{2}\)
Example 15.2.13.
Evaluate \(\ds \int_0^2\int_{y/2}^1 y\cos (x^3-1)\, dx\, dy\)
Solution.
Switching to a Type I integral, we have
\begin{equation*}
\int_0^1\int_0^{2x} y\cos(x^3-1)\, dy\, dx
\end{equation*}
which evaluates to \(\dfrac{2}{3}\sin (1)\)
Subsection 15.2.4 After Class Activities
Example 15.2.14.
Evaluate the following integrals. Sketch and label the region of integration.
- \(\ds \iint_D (y-3x)\, dA\text{,}\) where \(D = \lrbrace{(x,y)\mid 1\leq y\leq 2, -1\leq x\leq 1}\)
- \(\ds \iint y\sin x\, dA\text{,}\) where \(D\) is the region bounded by \(y = 0\text{,}\) \(y = x^2\text{,}\) and \(x= 1\text{.}\)
- \(\ds \iint xy\, dA\text{,}\) where \(D\) is the region enclosed by the quarter-circle \(y = \sqrt{1-x^2}\text{,}\) \(x\geq 0\text{,}\) and the coordinate axes.
Solution.
- \(\ds \iint_D (y-3x)\, dA\text{,}\) where \(D = \lrbrace{(x,y)\mid 1\leq y\leq 2, -1\leq x\leq 1}\)
- \(\ds \iint y\sin x\, dA\text{,}\) where \(D\) is the region bounded by \(y = 0\text{,}\) \(y = x^2\text{,}\) and \(x= 1\text{.}\)
- \(\ds \iint xy\, dA\text{,}\) where \(D\) is the region enclosed by the quarter-circle \(y = \sqrt{1-x^2}\text{,}\) \(x\geq 0\text{,}\) and the coordinate axes.
Example 15.2.15.
Evaluate the integral by reversing the order of integration. Sketch and label the region of integration.
- \(\displaystyle \ds \int_0^1\int_{3y}^3 e^{x^2}\, dx\, dy\)
- \(\displaystyle \ds \int_0^8 \int_{\sqrt[3]{y}}^2 e^{x^4}\, dx\, dy\)
Solution.
- \(\displaystyle \ds \int_0^1\int_{3y}^3 e^{x^2}\, dx\, dy\)
- \(\displaystyle \ds \int_0^8 \int_{\sqrt[3]{y}}^2 e^{x^4}\, dx\, dy\)
