Previously we have discussed methods of approximating the volume of a solid with regard to solids of known cross section and solids that are symmetric about an axis, via shells and washers. We are now discussing ways to approximate the volumes of more irregular solids—i.e. those without known cross section and or are not symmetric about an axis. We will first discuss the solid we will be approximating, followed by the method of approximation and calculating an actual volume by double integrals, and conclude with actual calculations of volume for the solid in question.
The solid that we will be considering in bounded on the \(xy\)-plane by the function \(y = \left| x \right| \) and \( y = 5 \); which results in the region shown below:
Vertically, or with respect to the \(z\)-axis, the solid is bounded by \( f(x,y) = 2y - x^2 \) which is shown in the following 3D image:
Methods
When approximating the volume of a solid of unknown cross section that is not symmetric about an axis one method is to divide the region in which the solid exists into a region of many rectangles, then by setting the height of the rectangle as an output of the function that bounds the top of our solid. In this way we can find the volume of this rectangular prism which in turn approximates solid's volume within that particular rectangular region. So, if we take one rectangle that totally encompasses the region where our solid lives we get a very rough approximation of its volume, more like an upper bound. For example, any solid object placed inside of a box must have volume less than or equal to that of the box. Now as we divide the region into more and more rectangles and subsequent rectangular prisms our approximation becomes more and more accurate as there is less "extra" volume in each prism that is not also in our solid; and if we increase these prisms infinitely we get an infinitely more accurate approximation. This infinitely more accurate approximation is actually the correct volume of said solid that is given by double integration as it essentially yields the volume of a very small prism, that could more precisely be referred to as a line, over every point that lies within the bounded region that makes up the base of our solid.
Another consideration is that if the height of these prisms are an output of the function that bounds the top of our solid, which actual point gives the output that we will use as our height? And simply you can use any point within the rectangle that forms the base of the rectangular prism as the one that gives the height. I have chosen the upper-righthand vertex for my approximation, and would recommend that you choose one of the four vertices, the center point, or the midpoint of one of the sides.
I decided to use 1x1 rectangles, which results in 30 rectangles of which 23 have heights, and thus volumes, that are of interest to use. The approximation of the solid is shown below:
Calculating the actual volume of our solids via integration requires the use of double integrals but is really rather simple. We define the region on the \(xy\)-plane within which our solid lives as \(R\) where \(x\) is bounded by \(a\) and \(b\) and \(y\) is bounded by \(g_1(x)\) and \(g_2(x)\) where \(g_1(x)\) is the lower-bound on \(R\) and \(g_2(x)\) is the upper-bound of \(R\) and \(f(x,y)\) is the function that bounds the top of our solid; or \[ R = { (x,y) \mid a \leq x \leq b, g_1(x) \leq y \leq g_2(x) } \\ V = \int_{R} \int f(x,y) dA = \int_{a}^b \int_{g_1(x)}^{g_2(x)} f(x,y) dy dx. \]
Calculations and Solutions
As stated above, I divided the region our solid lives in into 30 1x1 rectangles, of these 30 rectangles 23 of them had heights greater than zero, those with heights less than or equal to zero were ignored as they lay below or right on the plane and were thus not part of our volumes. Those rectangles and their sum, which is our approximate volume are shown in the table below:
Note that the \(z\) coordinates highlighted in green denote rectangles that actually contribute towards our approximations volume. As you can see from the table our approximation yielded a volume of 106 units\(^2\). Now we can see how good our approximation is by preforming the double integration— describes in the "Methods" section. This integration is setup can completed as follows: \[ V = \int_{-5}^5 \int_{\left| x \right|}^5 2y-x^2 dydx \\ = \int_{-5}^5 (y^2-x^2y) \mid_{\left| x \right|}^5 dx \\ \approx 62.5. \] Obviously, our approximation and volume via the integral are not that similar, in fact our actual volume from our double integration is only about 59% of our approximated volume. So, now let's see if using more rectangles gets us a more accurate approximate volume. These values and subsequent volume can be found in the table below, also with the green marking rectangles we care about:
Now our approximation has gone the other way and is actually smaller than our actual volume. This is due to volume that is lost when the upper-righthand vertex lies inside the top curve such that some volume is lost, but now our approximate volume is 85% of our actual volume which is a fairly precise approximation.
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