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You typed your integrand (x + y + x), Im assuming you meant (x + y + z).

This integration is a real pain. I'll go through the steps but skip all the algebraic simplifications in between.

This integration involves vartiable limits. All three limits start at zero since you have three bounding planes at zero. There is some flexibility in how you select the limits, so you can get the correct answer several ways. You need to make sure you do not include too much volume here.

First, look at the boundary plane where it intersects the xy plane. In the xy plane, z=0 so the boundary plane: x + 2y + 3z = 6 reduces to a line in xy coords: x + 2y = 6. Solving for y: y = 3 - x/2. This is the limit on the y coordinates so y will range from 0 to 3 - x/2.

The limit on x must include the region of the line in the positive xy quadrant. The relation y = 3 - x/2 gives the maximum x when y is minimum. Here that means y is zero so the maximum x is 6.

The final limit on z is just from zero to the bounding plane x + 2y + 3z = 6. Solving for z gives: z = 2 - x/3 - 2y/3

Now all the limits are complete, just integrate. Notice that the limits on z include x and y so this integration is done first. The limits on y include x so that is done second. Finally integrate on x.

Integrating on z between z=0 and z = 2 - x/3 - 2y/3 gives:

I (x + y + z) dx dy dz = ((x + y)z + (z^2)/2) dx dy

Applying limits and expanding:

(4x/3 - 5(x^2)/18 - 7xy/9 + 2y/3 - 4(y^2)/9 + 2) dx dy

Next integrate on y:

(4xy/3 - 5(x^2)y/18 - 7x(y^2)/18 + (y^2)/3 - 4(y^3)/27 + 2y) dx

Apply the limits of zero to 3 - x/2 and simplifying significantly:

(x/2 - 7(x^2)/12 + 13(x^3)/216 + 5 ) dx

Next integrate on x:

(x^2)/4 - 7(x^3)/36 + 13(x^4)/864 + 5x

Apply the limits of 0 to 6 gives 33/2

What a pain!

If you need more details, you can e-mail me.

Okay, no guarantees that I did this right. But maybe if we have the same answer, then we can both feel a little more confident. I assume we are calculating the volume of this solid. I also assume that you meant to take the triple integral of (x+y+z) dE.

This is where I might go wrong (It's been a while). Can we take dE=dxdydz? That's what I did. The limits of integration for x were 0 to 6, for y were 0 to 3, and for z were 0 to 2. 0 to 6 was obtained for x by plugging in y=0 and z=0 to see where these two planes intersect the line x+2y+3z=6. This is also how I found the limits for y and z.

Now, given that I'm correct up to now, the answer I come up with is that the volume is 198.

This was found by doing the following:

Int (from 0 to 2) Int (from 0 to 3) Int (from 0 to 6) [x+y+z] dxdydz

and just working from the inside out, starting with dx. Note that for dx, we treat y and z as constants. For example, the integral of ydx is xy.

Hope this helps.

Okay, I'll try it. I think your integral is (x+y+z) instead of (x+y+x). I sketched your plane, and since x = 6 - 2y - 3z is well-behaved (i.e., no fractions), we'll integrate on x first. After that, it probably doesn't matter, so we'll go in the order dx dy dz, inside-out.

Notation-wise, let's say int(a,b) f(x)dx means the integral from a to b.

Since we're integrating on x first, the projection of your plane onto the yz plane is 2y = 6 - 3z, or y = 3 - (3/2)z.

The triple integral is

int(0,2) int[0, 3-(3/2)z] int(0, 6-2y-3z) (x+y+z) dx dy dz

The first (inside) integral is

(1/2)x^2 + (y+z)x where x = 6-2y-3z (lower limit zero)

(1/2)x^2 = 2y^2 + 6(z-2)y + (9/2)z^2 - 18z + 18

(y+z)x = -2y^2 + (-5z+6)y - 3z^2 + 6z

(1/2)x^2 + (y+z)x = (z-6)y + (3/2)z^2 - 12z + 18

and the double integral becomes

int(0,2) int[0, 3-(3/2)z] [(z-6)y + (3/2)z^2 - 12z + 18] dy dz

The second integral is

(1/2)(z-6)y^2 + [(3/2)z^2 - 12z + 18]y where y = 3-(3/2)z

(lower limit zero)

(1/2)(z-6)y^2 = (9/8)z^3 - (45/4)z^2 + (63/2)z - 27

[(3/2)z^2 - 12z + 18]y = (-9/4)z^3 + (45/2)z^2 - 63z + 54

(1/2)(z-6)y^2 + [(3/2)z^2 - 12z + 18]y

= (-9/8)z^3 + (45/4)z^2 - (63/2)z + 27

(Interesting that the second term is negative two times the first term. There's probably a short-cut somewhere.)

The last integral becomes

int(0,2) [(-9/8)z^3 + (45/4)z^2 - (63/2)z + 27] dz

= (-9/32)z^4 + (15/4)z^3 - (63/4)z^2 + 27z where z=2

(lower limit zero)

= -9/2 + 30 - 63 + 54 = 33/2 (ANSWER)

And you were right. This was a lot of work. I hope your answer and mine match. I'm not going back to check my algebra!

this question has been up for a mutually as i won't be able to have faith no one else jumped on being the 1st to respond to it! i flow to do it even even with the undeniable fact that I stay in West Virginia, somewhat on the part of Morgantown and have hated Pitt all my existence yet we can all unite for Calculus. 2 0 3 0 = 6, x^2=4u^2 and the planar ellipse 9x^2<=36 is resembling the disk u^2+v^2<=a million The triple integrad of x^2dA=double int (4u^2)(6)dudv properly you combine it- which i think of you're able to desire to have finished top as much as that element 24[a million/2x+a million/4sin(2x)] (from 2pi to 0) * [a million/4r^4] (from a million to 0) ^ l l =a million no longer a million/2 (i think of you made your errors right here and subsequently the respond is 6pi