I do not see how I will be crossing the plane.

http://www.wolframalpha.com/input/?i=y%3D0+plane

Clearly I am staying on the plane when I go from (-1,0,0) to (1,0,0)

oh oops. my bad. You are probably correct then

V = I (x=-1 to 1, ...

this clearly crosses the plane x = 0 !

Definitely got confused with your y = 0 instead of x = 0.

Sorry for my previous post, too much drink last night I guess.

What I should have posted is :

V = I ( y=0 to 2, x=-1 to 1-y, (5-5x²)dxdy )

which is 20/3.



And what you did here is 100% correct. Same value : 20/3 (and probably a little easier to compute).

I only did it this way so you could see this pyramid has a base on both xy plane and xz plane so you can do the calculation either way and get to same result.

This post was edited by Xx Shin3d0wn xX on Nov 28 2014 08:53am

Great thank you everyone

For #3 I realized that the volume of a tetrahedron is 1/6 the volume of a parallelepiped with 3 converging sides so I was able to change the vertex (2,1,0) to (3,1,0) to make the tetrahedron easier but keep the volume the same.

With the new vertex I was left with only one plane that was not on the coordinate planes instead of two so the integrand and bounds of integration were much easier.

I did

V = Integral (x=0 to 3, y = 0 to x/3) (4/3)x - 4y dydx = 2

for question 1.
Take derivative of both equations wrt x and y. In the first part you showed that the system defines you and v implicitly as functions of x and y. therefore u = f(x,y) and v=g(x,y)
With respect to x
(5v^4 * v^2) + (x^5 * 2v * ∂v/∂x) + (2y^3 * ∂u/∂x) = 0
(3y * ∂u/∂x – uv^3) – (x * ∂u/∂x * v^3) – (xu * 3v^2 * ∂v/∂x) = 0
With respect to y
(x^5 * 2v * ∂v/∂y) + (6y^2 * u) + (2y^3 * ∂u/∂y) = 0
3u + (3y * ∂u/∂y) – (x * ∂u/∂y * u^3) – (xu * 3v^2 * ∂v/∂y) = 0
Evaluate each at (1,1,1,1)
5 + (2 * ∂v/∂x) + (2 * ∂u/∂x) = 0
(3 * ∂u/∂x) – 1 – (∂u/∂x) – (3 * ∂v/∂x) = 0
(2 * ∂v/∂y) + 6 + (2 * ∂u/∂y) = 0
3 + (3 * ∂u/∂y) – (∂u/∂y) – (3 * ∂v/∂y) = 0
Simplify and bring all the constants onto the right side
(2 * ∂v/∂x) + (2 * ∂u/∂x) = -5
(2 * ∂u/∂x) – (3 * ∂v/∂x) = 1
(2 * ∂v/∂y) + (2 * ∂u/∂y) = -6
(2 * ∂u/∂y) – (3 * ∂v/∂y) = -3
Hopefully now you can see that this system can be put into matrix form Ax = B
where A is the matrix with entries of the coefficients of your partial derivative, x is the matrix of partial derivatives and B is the matrix with the entries of the constants that are on the right side of your equations.
So to solve for x multiply B by the inverse of A
x = A^-1B
Won’t show that step because typing out matrices are a pain

For question 6, your method looks correct, however if the question had the planes x=0 AND y=0 feanur's first answer would be correct

Haha 'you'