This page covers double integrals in Cartesian coordinates on nonrectangular regions. This page continues the discussion from the double integrals on rectangular regions page.
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So, now we add a level of complexity where the region in the xyplane is not rectangular. This results in the limits of integration not being constant. There are only a couple of additional things that need to happen when setting up and evaluating double integrals with nonconstant limits.
First   It is imperative to plot the region in the plane when setting up the integral. 
Second   When switching the limits of integration, the limits themselves will change. 
Plotting The Region
As mentioned above, the double integral can be thought of as representing the volume of a solid above a region in the xyplane of height \(f(x,y)\) (the integrand). When working with nonrectangular regions, you need to plot the region in the xyplane. This will help you determine the limits of integration. Let us make this clear. This is not something that is optional that we are suggesting. This is a critical step in solving these problems and we have never found anyone that can do without it by doing it in their head. In fact, most instructors will require a plot for full credit on homework and exams. So don't neglect doing this in order to save time or energy or whatever. It's not worth it.
To help a bit and if your instructor allows it, you can use the free program winplot to plot the boundaries and then print it out and work with it for your homework. A link to the free program is on the tools page. We use winplot for many graphs on this site. But check with your instructor to make sure it is okay with them. This next panel explains how to describe the region once you have graphed it.
To describe an area in the xyplane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot (used to plot these graphs; we used gimp to add labels and other graphics). However, graphing by hand is usually the best and quickest way.
We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by \(f(x)\) (red line), \(g(x)\) (blue line) and \(x=a\) (black line).
[Remember that an equation like \(x=a\) can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]
Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).
Vertically
Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.
Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing \(g(x)\), no matter where we draw it. Similarly, the arrow always exits the area by crossing \(f(x)\), no matter where we draw it. Do you see that?
But wait, how far to the right does it go? We are not given that information. What we need to do is find the xvalue where the functions \(f(x)\) and \(g(x)\) intersect. You should be able to do that. We will call that point \((b,f(b))\). Also, we will call the left boundary \(x=a\). So now we have everything we need to describe this area. We give the final results below.
Vertical Arrow  

\( g(x) \leq y \leq f(x) \) 
arrow leaves through \(f(x)\) and enters through \(g(x)\) 
\( a \leq x \leq b \) 
arrow sweeps from left (\(x=a\)) to right (\(x=b\)) 
Horizontally
We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of \(f(x)\) and \(g(x)\) in terms of \(y\). ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations \(f(x) \to F(y)\) and \(g(x) \to G(y)\).
Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line \(x=a\). However, there is something strange going on at the point \((b,f(b))\). Notice that when the arrow is below \(f(b)\), the arrow exits through \(g(x)\) but when the arrow is above \(f(b)\), the arrow exits through \(f(x)\). This is a problem. To overcome this, we need to break the area into two parts at \(f(b)\).
Lower Section   This section is described by the arrow leaving through \(g(x)\). So the arrow sweeps from \(g(a)\) to \(g(b)\).
Upper Section   This section is described by the arrow leaving through \(f(x)\). The arrow sweeps from \(f(b)\) to \(f(a)\).
The total area is the combination of these two areas. The results are summarized below.
Horizontal Arrow  

lower section  
\( a \leq x \leq G(y) \) 
arrow leaves through \(G(y)\) and enters through \(x=a\) 
\( g(a) \leq y \leq g(b) \) 
arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\)) 
upper section  
\( a \leq x \leq F(y) \) 
arrow leaves through \(F(y)\) and enters through \(x=a\) 
\( f(b) \leq y \leq f(a) \) 
arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\)) 
Type 1 and Type 2 Regions
Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.
Here is a quick video clip going into more detail on Type 1 and Type 2 regions.
video by Krista King Math 

[ For some videos and practice problems dedicated to this topic, check out this page. ]
Reversing Order of Integration
So what do we do when the limits of integration are NOT constants and why would we want to switch the limits of integration anyway?
Let's answer the second question first. This is certainly a valid question since we don't just want to add extra work if we don't need to. And that's the key . . . do we really need to? The answer is yes. Sometimes the inside integral can not be evaluated, which means that the outside integral cannot be evaluated either. So, switching the order of integration will often allow us to evaluate the inside integral. You will see why soon.
When the area is not rectangular, the procedure is completely different for reversing the order of integration than we saw when the area was rectangular. The first critical step is to plot the region over which the integral is being evaluated. After plotting, you need to describe the region differently. It is time for a video to help you understand what you need to do.
Here is a great video that demonstrates, in general, how to switch the limits of integration for a nonrectangular region. Notice that he plots the region and how critical the plot is to his solution.
video by PatrickJMT 

A couple of things that he showed in that last video require comments.
First   Labeling the limits of integration showing which variable it was assigned to, is a powerful technique to keep track of the limits. Try to get in the habit of doing this ( if your instructor allows it, of course ), since later on ( with triple integrals ), it can help you keep track of what is going on.
Second   He drew an arrow to represent the limits of the inside limits which is extremely helpful. However, to represent the outside limits, it is better to think of the outside integral as sweeping that arrow across the region, which seems to sweep out the area itself.
Here is a video clip explaining some properties of double integrals.
video by Dr Chris Tisdell 

Okay, so now you have the basics. Time for some practice problems. After that, the next page you may want to go to explains using double integrals to calculate areas and volumes.
Practice
Unless otherwise instructed, evaluate these integrals, giving your answers in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.
Consider the triangular region R with vertices \((0, 0)\), \((4, 2)\) and \((6, 0)\). Set up both generic integrals \(\displaystyle{ \iint\limits_R { f(x,y) ~ dA } }\) over this region.
Problem Statement
Consider the triangular region R with vertices \((0, 0)\), \((4, 2)\) and \((6, 0)\). Set up both generic integrals \(\displaystyle{ \iint\limits_R { f(x,y) ~ dA } }\) over this region.
Solution
video by James Hamblin 

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Consider a region R bounded by \(y=x^3\) and \(y=4x\). Set up the two generic integrals \(\displaystyle{ \iint\limits_R { f(x,y) ~ dA } }\).
Problem Statement
Consider a region R bounded by \(y=x^3\) and \(y=4x\). Set up the two generic integrals \(\displaystyle{ \iint\limits_R { f(x,y) ~ dA } }\).
Solution
video by James Hamblin 

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\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{2y}{ 4 + 2x  y^2 ~dx } ~dy } }\)
Problem Statement 

\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{2y}{ 4 + 2x  y^2 ~dx } ~dy } }\)
Final Answer 

\( 29/6 \)
Problem Statement
\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{2y}{ 4 + 2x  y^2 ~dx } ~dy } }\)
Solution
video by The Organic Chemistry Tutor 

Final Answer
\( 29/6 \)
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\(\displaystyle{ \int_{0}^{4}{ \int_{\sqrt{x}}^{2} { \frac{x}{y^5+1} ~dy ~dx } } }\)
Problem Statement
\(\displaystyle{ \int_{0}^{4}{ \int_{\sqrt{x}}^{2} { \frac{x}{y^5+1} ~dy ~dx } } }\)
Solution
video by James Hamblin 

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Integrate \(\displaystyle{ \frac{2y}{(x^2+1)} }\) over the region \( \{ (x,y) ~~ 0 \leq x \leq 1,~0 \leq y \leq \sqrt{x} \} \)
Problem Statement
Integrate \(\displaystyle{ \frac{2y}{(x^2+1)} }\) over the region \( \{ (x,y) ~~ 0 \leq x \leq 1,~0 \leq y \leq \sqrt{x} \} \)
Solution
video by PatrickJMT 

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Integrate \( f(x,y) = x^2y+y^3 \) over the region \( \{(x,y) ~ \left ~ \right. x^2 + y^2 \leq 1, ~ x \geq 0, ~ y \geq 0 \} \)
Problem Statement
Integrate \( f(x,y) = x^2y+y^3 \) over the region \( \{(x,y) ~ \left ~ \right. x^2 + y^2 \leq 1, ~ x \geq 0, ~ y \geq 0 \} \)
Solution
video by Dr Chris Tisdell 

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Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ e^{x^2} dx } ~dy } }\)
Problem Statement
Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ e^{x^2} dx } ~dy } }\)
Solution
video by Dr Chris Tisdell 

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Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{x}^{1}{ \left( 1  y^2 \right)^{1/2} dy } ~dx } }\)
Problem Statement
Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{x}^{1}{ \left( 1  y^2 \right)^{1/2} dy } ~dx } }\)
Solution
video by Dr Chris Tisdell 

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\(\displaystyle{ \int_{0}^{1}{ \int_{ \sqrt{x}}^{1}{ \sqrt{1+y^3} ~dy } ~dx } }\)
Problem Statement
\(\displaystyle{ \int_{0}^{1}{ \int_{ \sqrt{x}}^{1}{ \sqrt{1+y^3} ~dy } ~dx } }\)
Solution
video by Dr Chris Tisdell 

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\(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ \sin(x^2) ~dx } ~dy } }\)
Problem Statement
\(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ \sin(x^2) ~dx } ~dy } }\)
Solution
video by Dr Chris Tisdell 

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\(\displaystyle{ \int_{0}^{1}{ \int_{y^2}^{1}{ 2\sqrt{x}~e^{x^2} ~dx } ~dy } }\)
Problem Statement
\(\displaystyle{ \int_{0}^{1}{ \int_{y^2}^{1}{ 2\sqrt{x}~e^{x^2} ~dx } ~dy } }\)
Solution
video by Dr Chris Tisdell 

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\(\displaystyle{ \int_{0}^{2}{ \int_{x^2}^{4}{ \frac{x^3}{\sqrt{x^4+y^2}} ~dy } ~dx } }\)
Problem Statement
\(\displaystyle{ \int_{0}^{2}{ \int_{x^2}^{4}{ \frac{x^3}{\sqrt{x^4+y^2}} ~dy } ~dx } }\)
Solution
video by Dr Chris Tisdell 

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\(\displaystyle{ \int_{2\pi}^0{ \int_{y^2}^{\cos(y^2)}{ 2y ~dx } ~dy } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int_{2\pi}^0{ \int_{y^2}^{\cos(y^2)}{ 2y ~dx } ~dy } }\)
Final Answer 

\( \sin(4\pi^2)  8\pi^4 \)
Problem Statement
Evaluate \(\displaystyle{ \int_{2\pi}^0{ \int_{y^2}^{\cos(y^2)}{ 2y ~dx } ~dy } }\)
Solution
video by World Wide Center of Mathematics 

Final Answer
\( \sin(4\pi^2)  8\pi^4 \)
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Practice Instructions
Unless otherwise instructed, evaluate these integrals, giving your answers in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.