# Thread: triple integral volume problem

1. ## triple integral volume problem

hi, i got some problems with solving a volume calculation, it looks like this:

Find the volume of the space that is created between the planes x=0, y=0; x+y+z=3 and x+2y=2
by the looks of it it seems fairly simple but i can't find a way to solve it, my main problem is to define the boundaries needed for the integral. anyone with some ideas?

2. Originally Posted by bjorn85
hi, i got some problems with solving a volume calculation, it looks like this:

Find the volume of the space that is created between the planes x=0, y=0; x+y+z=3 and x+2y=2
by the looks of it it seems fairly simple but i can't find a way to solve it, my main problem is to define the boundaries needed for the integral. anyone with some ideas?
Hi. As the problem is stated, the volume is infinite because there is no lower bound on $\displaystyle z.$

The constraints on $\displaystyle x,\ y$ can be written

$\displaystyle 0 \le x \le 2$
$\displaystyle 0 \le y \le 1-x/2.$

But for then for $\displaystyle z$ the constraint is

$\displaystyle ? \le z \le 3 - x - y.$

3. sorry, typo, it should be x=0, z=0, x+y+z=3 and x+2y=2

the problem is that the space is very angular(?)

4. The region is in the first quadrand created by the lines $\displaystyle x+y=3$ and $\displaystyle x+2y=2$
Since the "upper" curve is $\displaystyle z=3-x-y$ and the "lower" curve $\displaystyle z=0$ by Fubini's theorem
$\displaystyle \int \int_S \int 1dV=\int_A \int \int_0^{3-x-y}1 dz dA=$$\displaystyle \int_0^3\int_{-\frac{1}{2}x+1}^{-x+3}\int_0^{3-x-y} 1 dz dy dx 5. i've been calculating that and i get it to 63/24, the answer says 8/3, but should'nt x go from 0 -> 4 since the crossing between the upper and the lower line is on x = 4? 6. Let the master try. ------------------- 1)\displaystyle x+y+z=3, Take 3 non-colinear point I thing the bests are \displaystyle (3,0,0),(0,3,0),(0,0,3) And place them on the x-y-z graph and connect them. So you have a triangle. (In red) 2)\displaystyle x+2y=2, since this does not contain \displaystyle z you are going to do a regular two dimensional graph and them project is (move it staight) up and down to get the full plane. Take two points, say, \displaystyle (0,1), (2,0) (In blue). 3)Note the region, on the bottom, it was supplied by the diagram previously. 4)Now you can see what you are integrating. Thus, you have \displaystyle \int_A\int \int_{v(x,y)}^{u(x,y)} 1 dz dA But, the lower curve, \displaystyle v(x,y)=0 And, the upper curve, \displaystyle u(x,y)=3-x-y Now switch to the region \displaystyle A. It is a type I region thus, \displaystyle \int_a^b\int_{f(x)}^{g(x)}\int_0^{3-x-y}1dz\, dy\, dx Where,\displaystyle g(x) is upper curve which is when \displaystyle z=0 in the curve \displaystyle x+y+z=3 thus, \displaystyle x+y=3 thus, \displaystyle y=-x+3 And \displaystyle f(x) is lower cuve which is, \displaystyle x+2y=2 which is equivalent to, \displaystyle y=-\frac{1}{2}x+1 Thus, the volume integral thus far is, \displaystyle \int_a^b\int_{-\frac{1}{2}x+1}^{-x+3}\int_0^{3-x-y} 1\, dz\, dy\, dx Now, we move to find \displaystyle a,b this is the easiest not the region starts from \displaystyle x=0 and ends on \displaystyle x=3 Thus, we have (just like before), \displaystyle \int_0^3\int_{-\frac{1}{2}x+1}^{-x+3}\int_0^{3-x-y}1\, dz\, dy\, dx After the first integration, \displaystyle \left \int_0^3\int_{-\frac{1}{2}x+1}^{-x+3} z\right|^{3-x-y}_0 dy\, dx Thus, \displaystyle \int_0^3\int_{-\frac{1}{2}x+1}^{-x+3} 3-x-y\, dy\, dx Thus, \displaystyle \left \int_0^3 3y-xy-\frac{1}{2}y^2\right|^{-x+3}_{-\frac{1}{2}x+1} dx Thus, \displaystyle \int_0^3 3(-x+3)-x(-x+3)-\frac{1}{2}(-x+3)^2-3\left(-\frac{1}{2}x+1\right)$$\displaystyle +x\left(-\frac{1}{2}x+1\right)+\frac{1}{2}\left(-\frac{1}{2}x+1\right)^2dx$
Open and combine,
$\displaystyle \left \int_0^3 \frac{1}{8}x^2-x+5dx=\frac{1}{24}x^3-\frac{1}{2}x^2+5x\right|^3_0=\frac{93}{8}$

I think I did this correctly. Not sure why answeres do not match. Maybe, I understood this problem slightly diffrenctly.

7. Originally Posted by ThePerfectHacker
[snipped]

3)Note the region, on the bottom, it was supplied by the diagram previously.

[snipped]

Now, we move to find $\displaystyle a,b$ this is the easiest not the region starts from $\displaystyle x=0$ and ends on $\displaystyle x=3$

[snipped]

I think I did this correctly. Not sure why answeres do not match. Maybe, I understood this problem slightly diffrenctly.
If you change the upper bound on $\displaystyle x$ to 4, the iterated integral evaluates to 8/3 as it should. The diagram of (3) assumed $\displaystyle y \ge 0,$ which implied the upper bound for $\displaystyle x$ was 3. This was part of the problem first posted but not of the corrected problem.

Originally Posted by bjorn85
sorry, typo, it should be x=0, z=0, x+y+z=3 and x+2y=2.

8. yup thanks guys, got stucked there for a moment

9. Ah!, I see why our results differ, I did not consider the full region.