Double integral?

Question

If anyone could explain how the following is done step by step, it would be greatly appreciated!

Find the volume of the solid that lies below the surface z = f(x,y) and above the region in the xy-plane bounded by the given curves.

z = 9 - x - y; y = 0, x = 3, y = (2/3)x

Answer 1

First of all your question is not properly worded. I don't think z = f(x,y) has anything to do with this problem, in part because we don't know f(x,y). Unless z = 9 - x- y is what you mean by f(x,y).

Secondly those "curves" are really planes.

Thirdly it is extremely difficult to explain this without a diagram. But here goes.
The xy plane simply means z > 0
y = 0 means the xz plane, and for the given problem y > 0
x > 3 for the given problem.
z = 0 - x - y is a plane that is angled to the xy plane. It intersects the xy plane at the line x+y = 0 and cuts the z axis at z = 9.
y = 2/3 * x is another plane that is perpendicular to the xy plane.

To find the volume of the enclosed region, you actually need to perform a triple integral. The integration is not the tough part. It's finding the limits of integration.

What we need are intersections between the planes. Let's start off with defining 0 < z < 6
That's because at x = 3, the max value of z is 9-3-y = 6 for y = 0.
3 < x < 9-z
because the plane z = 9-x-y intersects the xz plane on the line z = 9-x or x = 9-z

Now to find the limits for y. Essentially y goes from 0 to whichever is "closer" between y = 2/3 x and y = 9-x-z.
2/3 x = 9-x-z when x = 3/5 (9-z)
So this x value where those two planes intersect varies with z.
When z = 4, this x value = 3 which is the minimum value of x.

So the limits are as follows:
0 < z < 4
3 < x < 3/5 (9-z) and 0 < y < 2/3 x
3/5 (9-z) < x < 9-z and 0 < y < 9-x-z

4 < z < 6
3 < x < 9-z and 0 < y < 9-x-z

So you have to integrate dx*dy*dz between those limits.

*EDIT*
The answerer below has a good approach, but makes the mistake of letting y go from 0 to 2/3 x. That is not the case. It only goes to 2/3 x when 3 < x < 5.4.
For 5.4 < x < 9, 0 < y < 9-x
You should get a final answer of
40.824 - 19 + 7.776 = 29.6

Answer 2

I'll try. And, contrary to what others may say, there is nothing wrong with the wording of the problem. It is perfectly understandable.

If you look at the curves in the xy-plane you will see that the area is bounded on the bottom by y=0 on the right by x=3 and on the top by the curve y=(2/3)x which goes from (0,0) to the point (3,2).

Form an element of area in the xy-plane that is dy by dx and then extend this upward to the function f to get an element of volume.

the element of volume is then (9 - x - y) dy dx

This must be integrated over the ranges of x and y to get the volume you want.

Let us do the y direction first and then the x. y is integrated from 0 to (2/3)x and then x goes from 0 to 3.

Integrate for y to get: 9y - xy - y^2/2 and evaluate from 0 to (2/3)x.
6x - (2/3)x^2 - (2/9)x^2 = 6x - (8/9)x^2

Now integrate in x and evaluate from 0 to 3.
3x^2 - (8/27)x^3 from x = 0 to x = 3
(27 - 8) - (0 -0) = 19

Check:
f = 9 at (0,0)
f = 6 at (3,0)
f = 4 at (3,2)
Volume = (1/2)(9 + 6 + 4)/3 = 19