What is the Method for Solving an Easy Triple Integral Problem?

[frozenguy]

Homework Statement

Use a triple integral to find the volume of the solid.
The solid in the first octant bounded by the coordinate planes and the plane 3x+6y+4z=12

Homework Equations

$$z=3-\frac{3}{4}x-\frac{3}{2}y$$

The Attempt at a Solution

So I'm using the triple integral type of dzdydx.

At first I set z and x to 0, then z and y to 0 to obtain x and y limits but I think that was incorrect. I'm trying to integrate 3x+6y+4z-12.

I guess the upper limits of z will be the above equation done in editor, z=3-3x/4-3y/2 and lower limit will be 0?

That integration gets really dirty really quick. I doubt it is right. Can someone help point me in the right direction?

 

[lanedance]

that sound like the way to go, why not try...

shouldn't get too bad as its just polynomials
$$\int_0^{?}dx \int_0^{?} dx\int_0^{3-\frac{3}{4}x-\frac{3}{2}y} dz$$

 

[HallsofIvy]

lanedance wrote that integral in a very pecular way- especially with two "dx"s! Clearly a typo.

I would write it as
$$\int_{x= 0}^a \int_{y=0}^{f(x)}\int_{z= 0}^{3- \frac{3}{4}x- \frac{3}{2}y}dz dy dx$$

That last integral, with respect to x, must give a numeric answer, of course, so the two limits must be numbers. What is the largest value x can take in this figure?

The second integral, with respect to y, can give a function of x. For every x, what is the largest value y can have? Projecting this figure into the z value is the same as taking z= 0. Since 3x+ 6y+ 4z= 12 becomes the line 3x+ 6y= 12 when z= 0: 6y= 12- 3x so what is y as a function of x?

Of course, for each (x, y), z goes from 0 up to the plane 3x+ 6y+ 4z= 12 which is what gives the formula for z that you give and the upper limit for the "z" integral.

 

[lanedance]

yep thanks Halls, meant
$$\int_0^{a}dx \int_0^{f(x)} dy\int_0^{3-\frac{3}{4}x-\frac{3}{2}y} dz$$

nice explanation, apologies if the notation confused things, I find it easier to keep track of the integrals in the operator notation.

Frozenguy, meant the same thing as Hall's post, but work with whatever you're comfortable with.

 

[frozenguy]

Ok so y=2-x/2 and that is the upper limit for dy?
So when it comes to that upper z limit getting squared I distributed the whole thing but now I'm thinking I didn't have to haha.

So is that the right track for the limits? and dx limits will be [0,4] correct?

 

[lanedance]

Ok so y=2-x/2 and that is the upper limit for dy?
So when it comes to that upper z limit getting squared I distributed the whole thing but now I'm thinking I didn't have to haha.

So is that the right track for the limits? and dx limits will be [0,4] correct?

sounds good - not sure why you were squaring things, but sounds like you're on the right track now

 

[frozenguy]

So can someone see any red flags? I'm getting a negative volume which doesn't make sense and plus the book says the answer is 4.

 

[lanedance]

your initial integrand should just be the volume element dV = dx.dy.dz = 1.dx.dy.dz

 

[lanedance]

In the form Hall's wrote:

$$\int \int \int dV = \int_{x= 0}^4 \int_{y=0}^{3-x/2}\int_{z= 0}^{3- \frac{3}{4}x- \frac{3}{2}y}1.dz dy dx$$

 

[frozenguy]

oh... Why is that? lol I'll try to read up on it but why don't we need to integrate a specific formula?

Thanks so much for your help

 

[lanedance]

no think of the integral as $\int_0^a f(x) dx$ adding up thr contribution from infintesimal itegrand (f(x) dx) over 0<x<a. If f(x) =1, $\int_0^a f(x) dx$ just adds up each incremental length dx to find the total 'x' length.

For the volume integral its the same, $\int \int \int dV = \int \int \int dx.dy.dz$. So as we want to calcaute the volume intergal think of it as adding up each infinitesimal block dV = dx.dy.dz, so the "function" to integrate must be identically 1.