Evaluating Triple Integral in 1st Octant: Q & 3x+2y+z=6

[triden]

Homework Statement

Evaluate \underbrace{\int\int\int}_{Q}(1-x) dzdydx

Where Q is the solid that lies in the first octant and below the plane:
3x + 2y + z = 6

The Attempt at a Solution

I guess my main problem is finding the integral limits. For dz, I arranged the equation of the plane so that z = 6-3x-2y. My limits for the dz integral would then be 0 to 6-3x-2y.

Not really sure if this is right or how to get the limits for dy and dx. Its been a year since I was in my multivariable class.

Chris

 

[LCKurtz]

That is correct for z limits. For x and y look at the triangle in the xy plane (z = 0).

 

[triden]

Ah...ok, I think I get it now. Just to confirm:

Since there is now a triangle in the x-y plane with a slope of (-3/2)x for the hypotenuse, my dy limits would be 0 to (-3/2)x and then my dx limits would be 0 to 2.

Chris

 

[LCKurtz]

Ah...ok, I think I get it now. Just to confirm:

Since there is now a triangle in the x-y plane with a slope of (-3/2)x for the hypotenuse, my dy limits would be 0 to (-3/2)x and then my dx limits would be 0 to 2.

Chris

Not quite. When you put z = 0, the equation of that slanted line is not y = -(3/2)x because it doesn't go through the origin. Just solve it for y...

 

[triden]

oops, yeah should have caught that. y = -(3/2)x + 3

Got an answer of 3 after a bit of algebra...seems about right!

Thanks a lot for your help, I appreciate it.