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Identity

Member

Registered: 2007-04-18

Posts: 934

Double integral

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)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Double integral

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..."

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Double integral

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