So here is the question. I get the width of the prism will be dx, but I am not sure how to go about finding A(y). If the formula is 1/2(B+b)h, and the lower base (b) is always 50, do I just use the fact that the other base relates to the height? Since h(tan48) = B/2 -25, do I just use that? Additionally, do need to use the table numbers to find a formula for x and y? Thanks in advance.
And no this is not my homework. I am trying to help explain it to someone else.
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Maybe you could make the question comprehensible.
I think you need to obtain the equation of the shape first.
First of all use the table of x and y to obtain an expression for the area of each cross section in terms of x. Integrate that wrt x to find the voume.
It sounds like they want you to use numerical methods to get the actual answer, so I’m assuming there isn’t a defined function y = f(x). You could plot it out and see if there is some parabola, ellipse or whatever that works.
But I think you’re on the right track. Just use basic geometry to write a formula for the area, A, as a function of x. Then integrate A over x from 0 to 600. You might just do a curve with an even numbered polynomial if there isn’t a function that exactly fits the data.
Yep, that’s true I assume there was some glaringly obvious best-fit function, but there isn’t.
Divide the volume into trapezoid prisms, work out the volume of each and add together.
Are you saying I am being unclear or that the question is?
You mean work out the volume between each set of given x and y values? It seems like there should be some easier way.
Can you work out the area A(x) of each trapezoid at each value of X? Then you can approximate the integral of A(x) dx using trapezoidal approximation to get the estimated volume.
If I was doing it that way, I’d interpolate the value of y at each midpoint and use that number for the height and multiply by dx to get the volume. A bit easier and probably a good enough approximation.
Kinda makes me wonder what kind of problem the author of the problem thinks this is. If you’re just supposed to figure the volume of a finite number of trapezoidal prisms and add them all up, then it hardly seems like a calculus problem at all.
If I tried to find the area under a curve by simply drawing some reasonable number of reasonably thin rectangles and adding up all their areas, would anyone say I was using calculus?
In a way, yeah. What you’ve just described is a Riemann sum, and a definite integral is defined as the limit of such Riemann sums as the “thinness of the rectangles” approaches zero. (At least, this is the most common definition encountered by Calculus students.)
In practice, definite integrals are evaluated either by (1) using the Fundamental Theorem of Calculus, which involves antidifferentiation, or (2) numerical methods, which are approximations (but can be made as close as you need) using Riemann sums like what you described, or refinements on this idea, like Simpson’s Rule or the Trapezoidal Rule.
The problem in question specifically states “Then evaluate the integral using a suitable numerical technique.”
Writing A as a function of just x is impossible, since we don’t have a relationship between y and x. The question also specifically states:
(Emphasis mine). Looking at the axes as represented in the figure, it’s clear that we want to integrate A dx from 0 to 600, where A is the area of each trapezium. If you write a formula for A in terms of y, you’ll have your integral. Note that the problem only asks you to “write down” the integral, not actually solve it. I’m avoiding writing down the solution because it’s pretty obvious, in the great tradition of “left as an exercise for the reader”.
IMHO, this isn’t a great problem, because it’s trying to do two things at once - emphasise the idea that there are a class of things we can calculate exactly via integration (such as the exact volume of earth), assuming we know the relationships between the variables, and at the same time getting you to calculate approximate values via numerical integration. I think it would have been better to give a formula for y in terms of x, and even if that integral can’t be computed exactly, the table of values provided can be used for numerical evaluation of the integral.