Introduction
You may now be familiar with known cross-section which takes the volume between curves. With this method we choose a certain shape to make the curve 3-dimensional to find the volume. But now let's say we want to find the volume of a function that is already 3-dimensional. Imagine we have a candle, and we want to find the volume of it. We would first need the function that gives us the "height" of the candle and we know this would be in terms of both x and y since the candle is 3-dimensional. We would then need the length and width of the base of the candle. Putting this together the height would be the function we integrate, and the width and length would be the bounds of each integral. We know the basics so let's look at an example. We custom make candles and had someone come in saying they wanted a slanted candle with the overall shape of a petal and we came up the function \[f(x,y)=xy+x\] and for the petal we used the functions: \begin{align*} f(x) &= \sqrt{x} \\ g(x)&= x^3 \end{align*} To make this candle we need to know how much wax we will need.
Compute Volume
We will first find the exact volume of this region. We will be using the double integral formula \[V =\int\int_{R}f(x,y)dA\] \(f(x,y)\) is the function we are finding, or the "height" as said above. The R are the bounds for the integrals respect to x and y. Applying this to our functions we would get \[V = \int_{0}^{1}\int_{x^3}^{\sqrt{x}}xy+x\ dydx\] The order of integration is dydx which means we will first integrate the function respect to y and then integrate respect to x. \begin{align*} V &= \int_{0}^{1}\int_{x^3}^{\sqrt{x}}xy+x\ dydx\ \\ &= \int_{0}^{1} \left. \frac{xy^2}{2}+xy\ \right\vert_{x^3}^{\sqrt{x}} dx \\ &= \int_{0}^{1} \left(\frac{x(\sqrt{x})^2}{2} + x(\sqrt{x})\right)\ -\ \left(\frac{x(x^3)^2}{2} + x(x^3)\right)\ dx \\ &= \int_{0}^{1} \left(\frac{x^2}{2}+x^\frac{3}{2}\right) - \left(\frac{x^7}{2} + x^4 \right)\ dx \\ &= \left.\left(\frac{x^3}{6}+\frac{2x^\frac{5}{2}}{5}\right)\ - \left(\frac{x^8}{16} + \frac{x^5}{5}\right)\right\vert_{0}^{1} \\ &= \left(\frac{1}{6}+\ \frac{2}{5}\right) - \left(\frac{1}{16}+\ \frac{1}{5}\right) \\ V&= \frac{73}{240} = 0.3041766 \end{align*} The volume is 0.3042 cubic units, that is the exact amount of wax that we would need to fill in the function \[f(x,y)= xy+x\] for the given bounds. To see how true this is we can also find the approximate value of this function with the same bounds.
In the picture to the left we can see the bounds f(x) (purple curve) and g(x) (red curve). To approximate this we would make squares or rectangles with the same length and width and to find the height we would use the f(x,y) function. We can see the f(x) and g(x) functions share points at (0,0) and (1,1). We will be making our rectangular prisms between these points. For this example, the length will be of 0.1 along the x-axis and the width would be 0.2 along the y-axis. This would make a total of 50 rectangular prisms. However, we know need to check if the rectangle is in the bounds for it to count towards the total volume. We can use any of the corners of the rectangle for the height, but the corner chosen must be the same for all the rectangles. Depending on which point we use the approximation will be an under or an over estimate, for the purpose of this example we will use the top right point of each rectangle. We can see at (0.1, 0.2) the point is inside the enclosed area which then means we would plug these points into the f(x,y) function to give us the height of 0.0024 for the rectangle. This would be done for all the rectangles giving us the table below
Using this process, we get an approximated volume of 0.334 cubic units. With the image on the right, we can see a visualization of how the approximations looked. Since we used the upper right point of each rectangle we over-estimated the total volume, but the approximation was still very close to the exact volume. This is what our slanted candle would look like with the shape of a petal.
Why these Functions
I chose to use this function because it was easy to integrate, but still gave us an idea of the entire process. In our end product, the candle, you can see how the bounds that were chosen affected the shape. We could have used more rectangular prisms which would have allowed the final shape to be more smooth. This would have also given us a more accurate answer. I chose to use these bounds because it showed that our inner integral can have bounds in terms of a variable. The dimensions of my model are 1 x 1 x 2 inches.
You may now be familiar with known cross-section which takes the volume between curves. With this method we choose a certain shape to make the curve 3-dimensional to find the volume. But now let's say we want to find the volume of a function that is already 3-dimensional. Imagine we have a candle, and we want to find the volume of it. We would first need the function that gives us the "height" of the candle and we know this would be in terms of both x and y since the candle is 3-dimensional. We would then need the length and width of the base of the candle. Putting this together the height would be the function we integrate, and the width and length would be the bounds of each integral. We know the basics so let's look at an example. We custom make candles and had someone come in saying they wanted a slanted candle with the overall shape of a petal and we came up the function \[f(x,y)=xy+x\] and for the petal we used the functions: \begin{align*} f(x) &= \sqrt{x} \\ g(x)&= x^3 \end{align*} To make this candle we need to know how much wax we will need.
Compute Volume
We will first find the exact volume of this region. We will be using the double integral formula \[V =\int\int_{R}f(x,y)dA\] \(f(x,y)\) is the function we are finding, or the "height" as said above. The R are the bounds for the integrals respect to x and y. Applying this to our functions we would get \[V = \int_{0}^{1}\int_{x^3}^{\sqrt{x}}xy+x\ dydx\] The order of integration is dydx which means we will first integrate the function respect to y and then integrate respect to x. \begin{align*} V &= \int_{0}^{1}\int_{x^3}^{\sqrt{x}}xy+x\ dydx\ \\ &= \int_{0}^{1} \left. \frac{xy^2}{2}+xy\ \right\vert_{x^3}^{\sqrt{x}} dx \\ &= \int_{0}^{1} \left(\frac{x(\sqrt{x})^2}{2} + x(\sqrt{x})\right)\ -\ \left(\frac{x(x^3)^2}{2} + x(x^3)\right)\ dx \\ &= \int_{0}^{1} \left(\frac{x^2}{2}+x^\frac{3}{2}\right) - \left(\frac{x^7}{2} + x^4 \right)\ dx \\ &= \left.\left(\frac{x^3}{6}+\frac{2x^\frac{5}{2}}{5}\right)\ - \left(\frac{x^8}{16} + \frac{x^5}{5}\right)\right\vert_{0}^{1} \\ &= \left(\frac{1}{6}+\ \frac{2}{5}\right) - \left(\frac{1}{16}+\ \frac{1}{5}\right) \\ V&= \frac{73}{240} = 0.3041766 \end{align*} The volume is 0.3042 cubic units, that is the exact amount of wax that we would need to fill in the function \[f(x,y)= xy+x\] for the given bounds. To see how true this is we can also find the approximate value of this function with the same bounds.
Using this process, we get an approximated volume of 0.334 cubic units. With the image on the right, we can see a visualization of how the approximations looked. Since we used the upper right point of each rectangle we over-estimated the total volume, but the approximation was still very close to the exact volume. This is what our slanted candle would look like with the shape of a petal.
Why these Functions
I chose to use this function because it was easy to integrate, but still gave us an idea of the entire process. In our end product, the candle, you can see how the bounds that were chosen affected the shape. We could have used more rectangular prisms which would have allowed the final shape to be more smooth. This would have also given us a more accurate answer. I chose to use these bounds because it showed that our inner integral can have bounds in terms of a variable. The dimensions of my model are 1 x 1 x 2 inches.
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