We are back again, once again, showing off a method of Integration from Calculus courses. Stepping beyond Calculus two, we will be going into one of the main ideas from a multiple variable calculus course, double integration. When doing a double integral, with two variables, we integrate over a region of $R^2$. The approximation is similar to what we would do for one variable, taking a double sum of heights times the area. In general, where we have a constant base area of $\Delta A$, we have the formula for approximation:
$V \approx \sum_{i=1}^{n} \sum_{j=1}^{m} \ f(x_i^*, y_j^*) \ \Delta A $
Following from the process from calculus one, by taking the limit of the double sum as i and j approach infinity, we get the exact volume.
The domain I decided to use for this is the curve sketched out by: $f(x,y) = x^2 + y^2$. This curve is a paraboloid, but that by itself is not very interesting. Instead of dealing with just this curve, we will be cutting out a region of a triangle, with the lines:
$y_1 = 5x$ and $y_2 = .2x$
Below is a graph of the region in Geogebra! The volume should end up looking like a chip you would use to eat salsa. Along with this, I wanted to continue my ideas of using curved shapes and triangles from all the other projects thus far.
Now, how do these integrals look? First, we need to talk about the approximations we want to use. Well, for our approximation, we will use rectangular prisms with square bases that are 1 by 1. The heights of each prism will be the value of the function $f(x,y) = x^2 + y^2$ at that given point. Along with this, our rectangles will be constructed using the top, right, end point of each square. Below is the dots we will be using, as well as the given domain of $x$ and $y$ we will use.
Now that we have our footing, we can discuss what the integrals would look like! The double integral is done by integrating over a specific Domain, giving us some volume. You make your bounds for $x$ and $y$ respectively by either: finding the values constrained for one variable, or find the equation of the line restricting one variable, written in terms of the other. For example, let's say you would looking at the region with lines: $y=x$ and $y=x^2$, from $0 \leq x \leq 2$. Your bounds for your $x$ values would be the given constraints on $x$ above, but your $y$ bounds would be from: $x^2$ to $x$, since we our restricting the $y$-values to be outputs of the function in terms of $x$.
This idea gives us the formulas for integration of two variables, where $D$ is the region which will give us the bounds for $x$ and $y$:
\[ \iint\limits_{D} f(x,y)\mathrm{\Delta}A \]
$\underline{Our \ Problem}$:
We want to approximate the volume of $f(x,y) = x^2 + y^2$, with the region being $y = 5x$ and $y=.2x$, along with $0 \leq x \leq 3$. Let us use the approximation first, and then see how it differs from the actual integration. Below is what the shape actually looks like on PrusaSlicer, along with the calculations used. By using the base area of the rectangle being 1, we find the approximate volume will be: $V_{approx} = 960$ units cubed. Now, what about the integration?
$\underline{Calculations \ for \ Approximation:}$
$\underline{Image \ of \ Shape:}$
Using the process discussed before, we want to solve the double integral that looks like:
$ V = \int_{0}^{3} \int_{.2x}^{5x}\ x^2 + y^2 dydx$
$\Rightarrow \int_{0}^{3} x^2*y + \frac{y^3}{3} dx \ |_{y=.2x}^{y=5x}$
$\Rightarrow \int_{0}^{3} 5x^3 + 125\frac{125x^3}{3} - .2x^3 - (.008)\frac{x^3}{3} dx$
From where, it is just integration methods learned in calculus one, which evaluates to:
$V = 940.896$ units cubed
The difference between our approximation and the actual volume is less than 20 units cubed, which makes sense due to us dropping some side cubes from not fitting exactly in our region using the top right point of the individual cube.
Overall, our approximation was very close to the actual volume of the object. I really enjoyed this due to my continued love of using triangles and rounded shapes from throughout the semester. As you can see in the picture of the 3D model, it did not appear as close to the chip I had envisioned, but is closer to some sort of a hook!
$V \approx \sum_{i=1}^{n} \sum_{j=1}^{m} \ f(x_i^*, y_j^*) \ \Delta A $
Following from the process from calculus one, by taking the limit of the double sum as i and j approach infinity, we get the exact volume.
The domain I decided to use for this is the curve sketched out by: $f(x,y) = x^2 + y^2$. This curve is a paraboloid, but that by itself is not very interesting. Instead of dealing with just this curve, we will be cutting out a region of a triangle, with the lines:
$y_1 = 5x$ and $y_2 = .2x$
Below is a graph of the region in Geogebra! The volume should end up looking like a chip you would use to eat salsa. Along with this, I wanted to continue my ideas of using curved shapes and triangles from all the other projects thus far.
Now, how do these integrals look? First, we need to talk about the approximations we want to use. Well, for our approximation, we will use rectangular prisms with square bases that are 1 by 1. The heights of each prism will be the value of the function $f(x,y) = x^2 + y^2$ at that given point. Along with this, our rectangles will be constructed using the top, right, end point of each square. Below is the dots we will be using, as well as the given domain of $x$ and $y$ we will use.
Now that we have our footing, we can discuss what the integrals would look like! The double integral is done by integrating over a specific Domain, giving us some volume. You make your bounds for $x$ and $y$ respectively by either: finding the values constrained for one variable, or find the equation of the line restricting one variable, written in terms of the other. For example, let's say you would looking at the region with lines: $y=x$ and $y=x^2$, from $0 \leq x \leq 2$. Your bounds for your $x$ values would be the given constraints on $x$ above, but your $y$ bounds would be from: $x^2$ to $x$, since we our restricting the $y$-values to be outputs of the function in terms of $x$.
This idea gives us the formulas for integration of two variables, where $D$ is the region which will give us the bounds for $x$ and $y$:
\[ \iint\limits_{D} f(x,y)\mathrm{\Delta}A \]
$\underline{Our \ Problem}$:
We want to approximate the volume of $f(x,y) = x^2 + y^2$, with the region being $y = 5x$ and $y=.2x$, along with $0 \leq x \leq 3$. Let us use the approximation first, and then see how it differs from the actual integration. Below is what the shape actually looks like on PrusaSlicer, along with the calculations used. By using the base area of the rectangle being 1, we find the approximate volume will be: $V_{approx} = 960$ units cubed. Now, what about the integration?
$\underline{Calculations \ for \ Approximation:}$
$\underline{Image \ of \ Shape:}$
Using the process discussed before, we want to solve the double integral that looks like:
$ V = \int_{0}^{3} \int_{.2x}^{5x}\ x^2 + y^2 dydx$
$\Rightarrow \int_{0}^{3} x^2*y + \frac{y^3}{3} dx \ |_{y=.2x}^{y=5x}$
$\Rightarrow \int_{0}^{3} 5x^3 + 125\frac{125x^3}{3} - .2x^3 - (.008)\frac{x^3}{3} dx$
From where, it is just integration methods learned in calculus one, which evaluates to:
$V = 940.896$ units cubed
The difference between our approximation and the actual volume is less than 20 units cubed, which makes sense due to us dropping some side cubes from not fitting exactly in our region using the top right point of the individual cube.
Overall, our approximation was very close to the actual volume of the object. I really enjoyed this due to my continued love of using triangles and rounded shapes from throughout the semester. As you can see in the picture of the 3D model, it did not appear as close to the chip I had envisioned, but is closer to some sort of a hook!
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