since the base we are integrating over is an ellipse with a = 2 and b = root2, we have converted to elliptic coordinates x = 2r cos(θ) and y = sqrt(2)r cos(θ)
and integrating the top function from my original post. when I convert it to elliptic coordinates it reduces to
(1 - r/2) r dr dθ...
ok so we have now got it to the point where we are integrating from z goes from the xy plane up to the cone surface, and the area of integration is the elliptical cylinder.
so in polar coordinates I think the integrand is 1 -r/2
but I am not sure what the limits of integration are for...
Homework Statement
find volume of the solid bounded by the surfaces
z = 1- \sqrt{\frac{x}{4}^2 + \frac{y}{2 sqrt{2}}^2}
and x^2/4 -x +(Y^2)/2 = 0
and the planes z = 0 and z = 1
Homework Equations
z = 1- \sqrt{\frac{x}{4}^2 + \frac{y}{2 sqrt{2}}^2}
and x^2/4 -x +(Y^2)/2 = 0...
Homework Statement
Evaluate the triple integral for the function \int\int\int y dV over that part of the cube 0 \leq x,y,z \leq 1 lying above the plane y +z = 1 and below the plane x+y+z = 2
Homework Equations
The Attempt at a Solution
This is the first attempt at a triple...