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Find the volume of the region bounded by the paraboloid z = x² + y² and below by the triangle enclosed by the line y = x, x = 0, and x + y = 2, in the xy-plane.
I'm having trouble setting the integrals up, can someone help thx
Last edited by Identity (2008-02-29 12:15:29)
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Start by drawing the lines in the x-y plane. Remember that these are 2-dimensional objects, and very easily seen. This is your "area of integration". Now pick x or y, you're choice. It will be easier or harder depending on which you pick, but you don't know which one it will be by just looking at it. If you wind up with a tough integral, try the other way.
Now that you've chosen what your favorite letter is (or you take a guess based upon intuition and experience), we first find two scalars that bound the function. I will choose x since it marks the spot.
Certainly the smallest x in this triangle is 0 (graph it... proof by "look!"). And I can say mostly from experience, but with a bit of logic and intuition, that the largest x will be where y = x and x+y = 2 intersect. That is, the point (a, b) where a = b and a + b = 2. This would mean that 2a = 2, and so a = 1 and b = 1. Therefore, my largest x is 1.
Now we look at the y's. I've already set up my bounds for x, I want to do so for y. What I need is some function that will give me precisely that triangle. So for example, on the far left when x = 0, my lower value is y = 0 (again, look at the picture). My upper value when x = 0 is given by x + y = 2, and since x = 0 I can use mathematica to solve it and get y = 2. Thus, for x = 0, I want to be integrating from 0 to 2.
But this doesn't mean I set up my inner integral from 0 to 2. Because at x = 1, my lower integral is going to be on the line y = x, and hence my lower is 1. And my upper bound for the integral defined by the line x + y = 2 is going to be 1 + y = 2, which of course is y = 1. So when x = 1, I'm going to be integrating from 1 to 1.
Things aren't very interesting at the end points, so lets look at when x = 1/2. Then my lower bound, y = x, is going to be y = 1/2. My upper bound, 1/2 + y = 2, is going to be 3/2. So at x = 1/2, I'm integrating from 1/2 to 3/2.
By now, it should be clear just from the way I'm talking that our upper and lower bounds of integration with respect to y are going to be a function that depends on x. And now the only question is what function will that be?
When x = 0: integrate from 0 to 2.
When x = 1/2: integrate from 1/2 to 3/2.
When x = 1: integrate from 1 to 1.
The bounds, as the graph clearly show, are in fact lines. So we can define a line through these three points (really only needed 2, I just used the third to give a non-endpoint example). And what do these lines turn out to be? Of course... the lines that we were given. The lower bound follows y = x and the upper, y = 2 - x.
Now we can set up our integral:
Look at that integral and think about it. When x = 0 (our outside integral), then the inside integral is from 0 to 1, exactly as we wanted it. When x = 1/2, our inside integral is from 1/2 to 3/2, again, exactly what we wanted. And finally, when x = 1, our inner integral is from 1 to 1.
Hopefully this is a clear way of explaining it. I was never taught the proper method for integrating in multiple dimensions, and I was shocked when I found out how simple it was.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Thanks ricky, did you have the axes so that x is vertical and y is horizontal? Because the way you explain it, it seems like that.
And the integral is then
?This is still complicated for me to visualize...
So
is the function giving the total area of the infinitisimal planes parallel to the y-axis of height f(x,y)...Then I graph this function, A(x) against x... and integrate from x = 0 to x = 1 (i like doing it this way i think)
Eventually I get Volume = 4/3 units cubed, right?
Thanks again
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