Thread: triple integral - two questions

question 1.
Find the volume of the tretrahedron bounded by the coordinate planes and the plane through (2,0,0), (0,3,0), and (0,0,1).

question 2.
Evaluate the triple integral of a function f(x,y,z) over the tetrahedron G with vertices (0,0,0), (1,1,0), (0,1,0) and (0,1,1).

Originally Posted by kittycat

question 1.
Find the volume of the tretrahedron bounded by the coordinate planes and the plane through (2,0,0), (0,3,0), and (0,0,1).

kitty, kitty...Remember how I said in the last problem that this was easy?

There is a very simple form you may not have learned along the way. It is called the Intercept Form. You will not believe how easy it is. It works for lines, too, but is particularly convenient for planes, given ALL the intercepts.

$\displaystyle \frac{x}{2} + \frac{y}{3} + \frac{z}{1} = 1$

Done.

You set up the integral.

Originally Posted by kittycat

question 2.
Evaluate the triple integral of a function f(x,y,z) over the tetrahedron G with vertices (0,0,0), (1,1,0), (0,1,0) and (0,1,1).

What would happen if you shifted the coordinate axes just a bit?

Try f(x,y-1,z) over (0,-1,0)(1,0,0)(0,0,0)(0,0,1). That looks an AWFUL lot like question 1, doesn't it? Give it a try.

HI TKHunny,

I don't get it! could you please explain to me more....especially for the first one. So what is the limits of the variables and why? Thank you very much.

I think to you that is easy because you are a master of the subject. But to me, the beginner, that is NOT EASY!

Originally Posted by kittycat

HI TKHunny,

I don't get it! could you please explain to me more....especially for the first one. So what is the limits of the variables and why? Thank you very much.

I think to you that is easy because you are a master of the subject. But to me, the beginner, that is NOT EASY!

Nothing easy about it. I am no master, by any account. Once you see the tetrahedron, you too will see how easy this particular type is. Again, once you've seen it.

Start with the equation of the plane. That's the hard part, in my mind. Once you have that, you're on your way.

It will look like this (I just picked an order.):

$\displaystyle \int_{0}^{}\int_{0}^{}\int_{0}^{}f(x,y,z)\;dx\;dy\ ;dz$

Solve for 'x' and plug it in.

$\displaystyle \int_{0}^{}\int_{0}^{}\int_{0}^{2(1-\frac{y}{3}-z)}f(x,y,z)\;dx\;dy\;dz$

We're done with 'x'. In a world where 'x' doesn't exist, solve the equation for 'y' and plug it in.

$\displaystyle \int_{0}^{}\int_{0}^{3(1-z)}\int_{0}^{2(1-\frac{y}{3}-z)}f(x,y,z)\;dx\;dy\;dz$

We're done with 'x' and 'y'. About what z domain are we caring?

$\displaystyle \int_{0}^{1}\int_{0}^{3(1-z)}\int_{0}^{2(1-\frac{y}{3}-z)}f(x,y,z)\;dx\;dy\;dz$

Just a piece at a time. No need to stress over the whole thing at once.

In this case, we just want the volume of the thing, so f(x,y,z) = 1.

Hi TKHunny,

Thank you very much.

Why do we need to set x=0 in order to find the y limit? Why don't we set y=0 to find the x limit?
To find the x, y limits in xy-plane, I usually use the projection of G on xy-plane ie. the region R.
I see you don't have to use this method. Why? Could you please teach me. Thank you very much.

Thank you very much.

Here's another way to find the equation of a plane with 3 known points.

Though, why bother, TKH's method is easiest. But.........................

Your points: $\displaystyle P_{1}=(2,0,0), \;\ P_{2}=(0,3,0); \;\ P_{3}=(0,0,1)$

$\displaystyle \overrightarrow{P_{1}P_{2}}=\langle{-2,3,0}\rangle$

$\displaystyle \overrightarrow{P_{1}P_{3}}=\langle{-2,0,1}\rangle$

Take the cross product of the two and get: $\displaystyle 3i+2j+6k$

Now, this is normal, so we can set up the equation using this and $\displaystyle P_{1}$

$\displaystyle 3(x-2)+2y+6z=0\Rightarrow{\boxed{3x+2y+6z-6=0}}$

Hi galactus,

Thank you very much. I used your way for my first attempt to do this question. I had the same equation of the plane as you given. But I don't know how to find the z , y , and x limits with this equation of the plane. Do you know how? Please teach me if you do. Thank you very muc.

Originally Posted by kittycat

question 1.
Find the volume of the tretrahedron bounded by the coordinate planes and the plane through (2,0,0), (0,3,0), and (0,0,1).

TKHunny did an excellent job on this question. I would have done it a much longer way though, since I don't recall seeing the "Intercept Form" before, my memory is horrible. here's is a longer way to do it, that is also more general, so we can use it given any set of points for a tetrahedron, even if all points are not intercepts, it is longer though.

recall how we define planes. one representation of a plane is:

$\displaystyle a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

where a,b, and c are the components of the normal vector, that is, $\displaystyle \bold { n } = <a,b,c>$ and (x_0,y_0,z_0) is a point the plane passes through.

We can find the normal vector using the cross-product (which we use to find a vector that is orthogonal to two given vectors).

Let A be the point on the x-axis, that is (2,0,0)
Let B be the point on the y-axis, that is (0,3,0)
Let C be the point on the z-axis, that is (0,0,1)

Then $\displaystyle \overrightarrow{AB} = <2,-3,0>$ and $\displaystyle \overrightarrow {AC} = <2,0,-1>$

Thus, $\displaystyle \bold { n } = \left| \begin{array}{ccc}\bold{i}&\bold{j}&\bold{k} \\ 2& {-3}& 0\\ 2&0&{-1} \end{array} \right|$

$\displaystyle = <3,2,6>$

Now using $\displaystyle \bold { n } = <3,2,6>$ and $\displaystyle (x_0, y_0, z_0) = A = (2,0,0)$, our plane is given by:

$\displaystyle P(x,y,z): 3(x - 2) + 2y + 6z = 0$

$\displaystyle \Rightarrow P(x,y,z): 3x + 2y + 6z = 6$, which is what TKHunny got.

Solving for z, we get:

$\displaystyle z = 1 - \frac {x}{2} - \frac {y}{3}$

In the xy-plane (set z = 0), we have the lines y = 0 and $\displaystyle y = 3 - \frac {3}{2}x$

On the x-axis, we have the limits being x = 0 to x = 2

(I hope you drew a diagram to know what I am talking about)

Now we set up our integral, we integrate the function 1 since we want to find the volume (as TKHunny rightly said), so we end up with.

$\displaystyle V = \int_{0}^{2} \int_{0}^{3 - 3x/2} \int_{0}^{1 - x/2 - y/3} dzdydx$

(we are bounded in the first octant by the coordinate planes, that's why all the lower limits are zero)

The integral looks a bit different from what TKHunny had, but it is the same thing. The difference is, I solved the plane for z rather than x (force of habit)

You can try this method for question 2 also if you wish

If you have any questions, please ask!


EDIT: galactus beat me to it

The same way TKH showed. Solve the equation for x to find the x limits.

x varies while y and z remain fixed.

Solving it for x we get $\displaystyle x=\frac{-2}{3}y-2z+2$.

Which is exactly what TKH had.

Set that to 0 and solve for y and get: $\displaystyle y=-3(z-1)$

This is the y-limit.

Hi Jhevon,

thank you very much.

I used the same method as you did for the first attempt to this question. I got the answer is -3 which is wrong. What answer do you get after all the integration?

I know TKHunny is correct because his answer 1 is the correct answer.

Originally Posted by kittycat

Hi Jhevon,

thank you very much.

I used the same method as you did for the first attempt to this question. I got the answer is -3 which is wrong. What answer do you get after all the integration?

I know TKHunny is correct because his answer 1 is the correct answer.

you evaluated incorrectly somehow. i got 1 as the answer as well. i'll admit, the algebra in my method is a bit harder that in TKHunny's method because you need to work with more fractions; but like i said, i wasn't looking for the easiest way, i was just going by the habit of solving for z first. you should be able to see how i got the limits that i did though, and apply it to TKHunny's method if the fractions are a problem for you

This is my 3th post!!!!!!!!!!

Hi Jhevon,

thank you very much. I'll double my work again.

Same as you , I think TKHunny's method is much easier

Originally Posted by kittycat

Hi Jhevon,

thank you very much. I'll double my work again.

Same as you , I think TKHunny's method is much easier

yup. my aim was to explain the methodology as to how we got the limits and the equation of the plane, did you get how to do those at least? after that, choosing the easiest order of integration is up to you

did you get the second one?

Hi Jhevon,

I got your way of working because I used exactly the same approach to solve this question but I got -3 as my answer. ( probably wrong calculation).

For TKHunny's method, I got it too. I am working on it at this moment. Made a mistake and didn't get 1. I am finding my mistakes in calculation.