Bounds for Triple Integral Region over Triangle

[Kuma]

Homework Statement

evaluate the integral:

int B of z DV where B is the region between the planes: z = x+y, z = 3x+5y and lies over the triangle with vertices (0,0), (0,1), (1,0)

Homework Equations

The Attempt at a Solution

I'm having some trouble trying to figure out the bounds of the 3d region. I can visualize what it looks like and draw it out by plugging in the points of the triangle into each plane and just drawing the region in between the points.

 

[Mark44]

Homework Statement

evaluate the integral:

int B of z DV where B is the region between the planes: z = x+y, z = 3x+5y and lies over the triangle with vertices (0,0), (0,1), (1,0)

Homework Equations

The Attempt at a Solution

I'm having some trouble trying to figure out the bounds of the 3d region. I can visualize what it looks like and draw it out by plugging in the points of the triangle into each plane and just drawing the region in between the points.

The two planes, z = x + y and z = 3x + 5y, both go through the origin, which is the only point in common between the two planes and the triangle in the xy plane.

The only slightly complicated part that I see is coming up with a description for the triangular region.

 

[Kuma]

I'm new to triple integrals and figuring out the bounds for each variable is the most confusing.

the first one, for x i said goes from 0 to the line y-1. Not sure if that's right. Would z go from 3x+5y to x+y? Since its the region between the planes. But I can't figure out a bound for y.

 

[Mark44]

I'm new to triple integrals and figuring out the bounds for each variable is the most confusing.

the first one, for x i said goes from 0 to the line y-1. Not sure if that's right.

If you integrate across first, x ranges from x = 0 to x = 1 - y. Here you are using thin horizontal strips.

To pick up all of the horizontal strips, the strips range from y = _ to y = _? (Fill in the blanks.)

Would z go from 3x+5y to x+y?

You need to go from the lower plane to the upper plane. The plane z = x + y is the lower one.

Since its the region between the planes. But I can't figure out a bound for y.

 

[flyingpig]

I haven't read this in detail, but shouldn't you see what the planes look like on xy-plane combined with the triangle?

 

[Kuma]

Y would go from 0 to 1?

 

[Mark44]

Y would go from 0 to 1?

Yes.

 

[Kuma]

Does the order of integration matter? I'm getting an answer in terms of x and y if i do it with the order dx dy dz. Shouldn't it be a number?

 

[Mark44]

Yes, of course the order matters. If you are getting something other than a number, you're doing something wrong.

 

[Kuma]

So how do i determine how the order should go?

 

[Mark44]

I would integrate in this order: z, x, y. You could also do it in the order z, y, x, but you would have to change your integration limits slightly.

 

[Kuma]

Alright. But how'd you determine in which way the order should go? Is there a rule?

 

[jackmell]

You've got to approach this more methodical. Can you just draw the 3D coordinate axes with z going up, y going into the plane of the paper, x going across? Just that much. Ok, when you got a choice, try and integrate the most natural way:

\iiint dzdydx

Now if you want, read this one:

https://www.physicsforums.com/showthread.php?t=554329

Alright, if you did, then you'll know the 1-2-3 rule alright? x is the outer integral and it goes from point a to point b. y is the center integral and it goes from curve g(x) to curve h(x). The inner one then goes from surface f(x,y) to surface p(x,y) so write:

\int_a^b \int_{g(x)}^{h(x)}\int_{f(x,y)}^{p(x,y)} dzdydx

Ok, now just for now scrap the inner integral and just look at the area to be integrate over:

\int_a^b \int_{g(x)}^{h(x)}dydx

Draw that triangle (in the x-y plane) over that nice plot of the 3D coordinate axes you made. Look at it carefully. Now, what must a and b be for x and what are g(x) and h(x) for y? Just get that part straight now.