# Double Integral

• Apr 7th 2005, 03:09 PM
Jojo B
Double Integral
A space is bounded by x = 0, y = 0, xy-plane, and the plane: 3x + 2y + z = 6. Find the volume using a double integral.
• Apr 8th 2005, 10:57 AM
billh
First find the region in the xy plane where the function intersects the x and y axes. Looks like a triangle in this case. Then set the interval for the outer integral, say x from 0 to the x-intercept. Then the interval for the inner integral is from y=0 to y as a function of x. Find this function by setting z to zero and solving for y. Then evaluate the partial integrals starting from the inner integral for y. For fun, switch the order of integration, as is allowed by Fubini's theorem.

Hey, we just went over all this in my Calc class so it is fresh in mind!
• Apr 9th 2005, 08:35 PM
ticbol
x=0 is the yz plane.
y=0 is the xz plane.

Since in the plane 3x +2y +z = 6
the z has a coefficient of unity or 1, then we use the the z-coordinate as the height of the dV,
and the dA will be on the xy plane, or dA = dx*dy.

dV = z*dy*dx

The boundaries of dy are from y=0 to y of the line 3x +2y = 6
2y = 6 -3x
y = (6 -3x)2 = 3 -1.5x
Or, dy is [0,(3 -1.5x)]

The boundaries of dx then are from x=0 to x=2
Or, dx is [0,2]

The z.
3x +2y +z = 6
z = 6 -3x -2y

So,
dV = (6 -3x -2y)dy*dx
Then,
V = INT.(0-->2){INT.(0-->3 -1.5x)[6 -3x -2y]dy} dx
V = INT.(0-->2){[6y -(3x)y -(2y^2)/2](0-->3 -1.5x)} dx
V = INT.(0-->2){[6(3 -1.5x) -(3x)(3 -1.5x) -(3 -1.5x)^2] -[0]} dx
V = INT.(0-->2){18 -9x -(9x -4.5x^2) -6y -(9 -9x +2.25x^2)} dx
V = INT.(0-->2){18 -9x -9x +4.5x^2 -6y -9 +9x -2.25x^2} dx
V = INT.(0-->2){2.25x^2 -9x +9} dx
V = [(2.25x^3)/3 -(9x^2)/2 +9x](0-->2)
V = [(2.25 *8)/3 -(9 *4)/2 +9*2] -[0]
V = 6 -18 +18