Area Between Curves in Which One of the Boundaries is Itself a Function

Factual Questions
#1

I’ll preface this by saying that this is not a homework problem. It’s more or less a geometry problem that I’m attempting to solve for a relative of mine regarding acreage.

There’s a plot of land with the following dimensions: 494.5’ x 105’ x 438.71’ x 105’. The shape is roughly like this (slope of bottom side greatly exaggerated):

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‘’’/
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Based on some other dimensions I’ve been able to determine functions that model each side, assuming the origin is placed at the top left. The top side is simply y=0 from x=0 to 105.6. The left side is y=(-0.155)*x from x =0 to 75.743. The bottom side is y=(0.568)*x - 531.69 from x=75.74 to 172.798. The right side is y=(-0.155)*x-16.368 from x=1-5.6 to 172.798.

Clearly, there are several ways I could solve this. Since the top and bottom are nearly equal, I could approximate the shape with a trapezoid and find the area that way. I could also divide this into many triangles and solve it that way. However, one of my quirks is that I get fixated on solving a problem a particular way and refuse to be satisfied with an answer obtained any other way. I’d like to solve this with integral calculus. If the left and right curves weren’t parallel (i.e. they intersected at a point), I could find the area between them easily. But the fact that they’re parallel and one of the bounding curves is not constant is throwing me quite a roadblock. Frankly, I’m embarrassed to admit I can’t solve this with calculus.

Any tips?

#2

It’s going to be a lot easier if do your equations to have the bottom left be at (0,0). At that point, you can use calculus to get the area. It will just be the integral of F(top)-F(bottom) over the length of the area in question.

#3

The left & right sides being parallel makes things a bit simpler - you can rotate the shape to make it a bit easier to calculate.

If you didn’t want to do that, or were dealing with, say a concave shape with uneven sides, the general way to do it using integral calculus is to split it up into pieces that you can solve.

Another tip - any spaces get stripped out when the board displays your post. You need to put it in a code block to keep them. Though it still doesn’t come out quite right, since it will use a fixed width font - trial and error with preview is your best bet then.

For example, here’s what you posted, when put in a code block



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#4

A double integral would work, as well, where one of the limits of integration was a function.