Expert: Alon Mandes - 9/3/2008

Question
QUESTION: Hi for my summer math packet there is a question that I am not sure how to answer. It ask to setup the triple integral that represent the volume of a tetrahedron that is enclosed by the coordinate planes and the plane 2x y z=4. I am not confused and not sure how to start this problem.

ANSWER: The volume of the three-dimensional region E is given by the integral ∫∫∫dv=∫∫∫dxdydz. We need to set the boundaries of each
variable. Let's take a look at our plane 2x+y+z=4, it can be written
as z=4-2x-y. So, we can claim that z varies from 0 to 4-2x-y. Now
we need to find the variety of x&y. To do so, let's find out what
shade or region this plane cover from the x-y plain :
In the x-y plane, z=0, meaning : 2x+y=4 --> y=4-2x. This is the equation of the line which represent the limit of the region. (note that the other limits are the lines x=0 & y=0). Now we know that y
varies from 0 to 4-2x. Good, but what about the boundaries of x ??
Well, when y=0 & z=0, x=2 (because 2x+0+0=4) So we can say that x
varies from 0 to 2. Let's sum up our calculations :
1. dz : 0 --> 4-2x-y
2. dy : 0 --> 4-2x
3. dx : 0 --> 3
Now we will move to calculating the triple integral :
∫∫∫dxdydz==∫∫[4-2x-y]dxdy=∫dx[4y-2xy-(y^2)/2]{y:0-->4-2x}=
=∫[4(4-2x)-2x(4-2x)-((4-2x)^2)/2]dx { x goes from 0 to 2 }
I'll leave it to you as an exercise to calculate this immediate
integral.

Alon.



---------- FOLLOW-UP ----------

QUESTION: Hi thanks that was a great help, but I had one more question for you. For another problem it asks given f(x,y,z)= (x^2)+yz, x= prcos(theta), and z= p+r, find the partial of f with respect to p, the partial of f with respect to r, and the partyial of f with respect to theta when p=2, r=3 and theta=0. Again thanks for your help.

ANSWER: If you mean by partial, partial derivative , then to calculate partial derivative , I need to know y(p,r,theta).

Alon.

---------- FOLLOW-UP ----------

QUESTION: p=2 r=3 and theta =0

Answer
The chain rule of partial derivative is
df/dr=(df/dx)(dx/dr)+(df/dy)(dy/dr)+(df/dz)(dz/dr), the same as
for p & theta.
For example df/dr=2x(pcos[theta])+z*1=2rpcos(theta)+p+r+y
df/dr|point=2*3*1+2+3+y=11+y
You may proceed to find df/dp & df/d(theta)  

Note that y is missing because you didn't give me his form as function of p,r & theta !

Alon.