How do we evaluate a double integral over a general region which is non-rectangular?
In other words, how do we evaluate a region over a two-dimensional area without fixed values but over a simple, closed curve instead.
This means the regions are being divided into groups according to their boundaries, and the method of evaluation is known as double integrals over general regions (non-rectangular areas).
Let’s take a closer look at what this means…
Recall how we found the area under a curve in single variable calculus, where we located the “top” and “bottom” functions for a specified interval using vertical or horizontal slices?
Well, we will do the same, just with double integrals!
Double Integrals Over A Region
Alright, so we are asked to find the volume under the surface by integrating over a two-dimensional area.
How can we divide the region? Two ways.
Type 1 Regions Are When We Slice The Two-Dimensional Region Vertically
Thus, creating a “dydx” double integral, where we first integrate \(f\left( {x,y} \right)\) with respect to \(y\), keeping \(x\) fixed. And then integrating the result with respect to \(x\) from \(a\) to \(b\).
\begin{equation}
\int_{x=a}^{x=b} \int_{y=g_{1}(x)}^{y=g_{2}(x)} f(x, y) d y d x=\int_{a}^{b}\left(\int_{g_{1}(x)}^{g_{2}(x)} f(x, y) d y\right) d x
\end{equation}
Type 2 Regions Are When We Slice The Two-Dimensional Region Horizontally
This creates a “dxdy” double integral, where we first integrate \(f\left( {x,y} \right)\) with respect to \(x\), keeping \(y\) fixed. And then integrating the result with respect to \(y\) from \(c\) to \(d\).
\begin{equation}
\int_{y=c}^{y=d} \int_{x=h_{1}(x)}^{x=h_{2}(x)} f(x, y) d x d y=\int_{c}^{d}\left(\int_{h_{1}(x)}^{h_{2}(x)} f(x, y) d x\right) d y
\end{equation}
So, all we have to do is find our top and bottom functions for our region and integrate!
Example
Let’s look at an example.
Evaluate \(\iint_{R}\left(4 x+y^{3}\right) d A\) two ways where R is the region enclosed by the graphs \(y = \sqrt x \) and \(y = \frac{1}{2}x\).
Okay, so the first thing we will do is graph our bounded region \(R\).
Now it’s time to define our region by slicing vertically and constructing a Type 1 double integral.
Our top function is \(y = \sqrt x \) and our bottom function is \(y = \frac{1}{2}x\), and our width is from \(\left[ {0,4} \right]\) along the x-axis, such that \(\frac{1}{2}x \le y \le \sqrt x \) and \(0 \le x \le 4\).
So, the integral becomes:
\begin{equation}
\int_{x=0}^{x=4} \int_{y=\frac{1}{2} x}^{y=\sqrt{x}}\left(4 x+y^{3}\right) d y d x=\int_{0}^{4}\left(\int_{\frac{1}{2} x}^{\sqrt{x}}\left(4 x+y^{3}\right) d y\right) d x=\frac{32}{3}
\end{equation}Next, we will take our same region but slice it horizontally and construct a Type 2 double integral.
This time we notice that our top function is \(x = {y^2}\) and our bottom function is \(x = 2y\), and our width is from \(\left[ {0,2} \right]\) along the y-axis, such that \({y^2} \le x \le 2y\) and \(0 \le y \le 2\).
Now we evaluate the integrals:
\begin{equation}
\int_{y=0}^{y=2} \int_{x=y^{2}}^{x=2y}\left(4 x+y^{3}\right) d x d y=\int_{0}^{2}\left(\int_{y^{2}}^{2 y}\left(4 x+y^{3}\right) d x\right) d y=\frac{32}{3}
\end{equation}And just as we would expect, both vertical and horizontal slices yield the same answer.
Not bad, right?
But there may be a tiny hiccup that can occur.
Sometimes we are unable to evaluate a double integral in its given state.
What do we do?
We use the power of Fubini’s Theorem, that we learned in our previous lesson for evaluating double integrals over rectangles and reverse the order of integration.
Now, this comes with a warning, as it’s not as easy to switch the order of integration when the limits on the integrals are variables and not numerals. Therefore, we will need to pay close attention to our two-dimensional region and go slowly.
Don’t worry. We’ll go through the steps needed to switch the order of integration in great detail within our lesson, all while tackling some pretty interesting questions.
It’s going to be fun learning how to apply double Integrals over non-rectangles (i.e., general regions), so let’s get to it!
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