Integration for Over Regions in the Plane

Introduction
From the previous weeks, the topics and concepts that we have learned have built on top of one another. Using the information we have learned thus far, we are now ready to dive deeper into the calculus world. Integration has been straight forward by integrating functions just on the two-dimensional world. What if we added the third-dimension?! Can integration still be done? The answer is yes, and using integration in the third-dimension is slightly more difficult, but can still be understood quite easily. Today, we will be using double integrals to calculate the volume of a region between two surfaces.
Double Integrals?
As intimidating as it may sound, double integrals are very useful and easy to learn. To compute a double integral, we just have to evaluate the double integral over a region that is either bounded by two horizontal and two functions of x or two vertical lines and two functions of y. Although it may seem difficult, a majority of the process is just setting up the integrals. Lets look at an example, so we can further understand how to set up and evaluate double integrals. Today, I chose the function \(f(x,y)=2yx^2+9y^3\). We are going to set \(D\) as the region bounded by \(f(x)=\frac{1}{4}y^2\) and \(g(x)=\frac{3}{2}y\). Lets take a second to visualize the region that we will be integrating.

Making a sketch of the region is important, for it allows us to not only visualize the region, but also to determine the order of integration. However, this example is nice and will allow for the integration to be done in either order. I have already chose the regions \(f(x)=\frac{1}{4}y^2\) and \(g(x)=\frac{3}{2}y\) which will make us integrate with respects to y first then x. Because of this, our bounds of integration will be \(\frac{2}{3}x\leq y\leq 2\sqrt{x}\) and \(0\leq x\leq 9\). Now that we know the order of our integration and the bounds for the region, we can both approximate and compute the volume of the region.
Approximating and Computing the Double Integration
Computing the volume would be easy at this point since we have figured out all the setup, but how would we approximate the volume of the region? Revist the graph from above. We are going to approximate the volume of the region by dividing the entire region into rectangular prisms. Each rectangular prism will have the dimensions 0.5 by 0.5 and a height that corresponds with the value of \(f(x,y)\) at that point. For this example, we are going to count the rectangular prisms that their top-right corner lies within the region. This means that any rectangular prism that lies outside the region will not have its volume included in the approximation. See the image that shows which rectangular prisms would be included.

Below is a table that includes the approximation of the region by taking the volume of the rectangular prisms that have their top-right corner within the region.

From this data, we are able to approximate the volume of the region. Our approximation gave us a \(Volume = 5569.6 units^3\). Now that we have approximated the volume of our region, we can compute the actual volume through the integration we discussed earlier. Lets revisit the information that we gathered earlier. We know that we will be integrating the function \(f(x,y)=2yx^2+9y^3\) that is bounded by \(f(x)=\frac{1}{4}y^2\) and \(g(x)=\frac{3}{2}y\). Lets set up the integration. First, the main formula for integration for over regions in the plane is the following: \[\iint\limits_{D} f(x,y)dA\] Now, if we subsitute in the function and the bounds we get the following integral: \[\iint\limits_{D}{{f(x,y)dA}} = \int_{0}^{9}{{\int_{{\frac{2}{3}x}}^{{2\sqrt x }}{{2yx^2+9y^3\,dy}}\,dx}}\] Notice how I chose to integrate with respect to y first. For this example, integrating with respect to x or y first does not matter, other examples may, so make sure you sketch your region to identify this! Next, we will finish the integration to compute the exact volume of our region. \begin{align*}\iint\limits_{D}{{2y{x^2} + 9{y^3}\,dA}} & = \int_{0}^{9}{{\int_{{\frac{2}{3}x}}^{{2\sqrt x }}{{2y{x^2} + 9{y^3}\,dy}}\,dx}}\\ & = \int_{0}^{9}{{\left. {\left( {{y^2}{x^2} + \frac{9}{4}{y^4}} \right)} \right|_{\frac{2}{3}x}^{2\sqrt x }\,dx}}\\ & = \int_{0}^{9}{{4x\left( {{x^2}} \right) + \frac{9}{4}\left( {16{x^2}} \right) - \left[ {\frac{4}{9}{x^2}\left( {{x^2}} \right) + \frac{9}{4}\left( {\frac{{16}}{{81}}{x^4}} \right)} \right]\,dx}}\\ & = \int_{0}^{9}{{36{x^2} + 4{x^3} - \frac{8}{9}{x^4}\,dx}}\end{align*} \begin{align*}= \left. {\left( {12{x^3} + {x^4} - \frac{8}{{45}}{x^5}} \right)} \right|_0^9 = {{\frac{{24057}}{5} = 4811.4units^3}}\end{align*} From our integration, we can see that we were off by quite a bit with our approximation. However, this is why it is imperative to learn how to use double integrals, especially when the numbers are significant to a real world situation! The image below shows the region with the corresponding approximated volume for its respective rectangular prism. As you can see, the object is quite large, thus when I actually print my object, it will need to be scaled down significantly. It is currently 9 x 6 x 2916 inches, but it has already been scaled down.

Why These Functions?
The function and the corresponding regions of integration were chosen for a few reason. For the obvious reason, it was a nice function to integrate. There weren't any crazy trig functions, inverses, etc. Sometimes it does matter in what order you integrate and having a function in which it does not matter what order you integrate allows for someone who is just learning the topic to start off on an easier function. This will allow for the person to create a foundation of the topic. Lastly, the bounds allowed for the creation of a half flower petal which relates to the function I chose for our first post on washers.