Evaluating Integrals on a Cone & Plane Intersection

[earn1087]

Homework Statement

let c be the curve of intersection of the cone z= sqrt(x^2+y^2) and the plane 3z= y+4, taken once anticlockwise when viewed from above.

(i) evaluate
∫c (sinx - y)dx +(x+cosx)dy + (e^z + z)dz

(ii) let s be the surface of the cone z= sqrt(x^2+y^2) below the plane 3z= y+4 and above the xy plane.

let F = (sinx - y)i +(x+cosx)j + (e^z + z)k

evaluate
∫∫s [(curlF) . nds], where n points downwards .

Homework Equations

The Attempt at a Solution

for(i) i attempted to create parameters however it creates a messy equation to integrate. i also don't know if i used the parameters correctly.

this is not homework, its the final question of a revision paper with no solutions

 

[mighty2000]

Thank you for your question. I am a scientist and I would be happy to help you with this problem.

First, let's take a look at the curve c. We can see that it is the intersection of the cone z=sqrt(x^2+y^2) and the plane 3z=y+4. We can rewrite the equation of the plane as z=(1/3)y+(4/3). Now, we can substitute this into the equation of the cone to get z=sqrt(x^2+y^2)=(1/3)y+(4/3). We can rearrange this to get y=3z-4, which we can then substitute into the equation of the cone to get z=sqrt(x^2+(3z-4)^2). This is the equation of the curve c.

Now, we can use these equations to evaluate the integral. First, let's find the limits of integration. We can see that the curve c is a circle, so we can use polar coordinates to integrate. The limits for r would be 0 to 3z-4, and the limits for θ would be 0 to 2π. Now, let's rewrite the integral using these parameters:

∫c (sinx - y)dx +(x+cosx)dy + (e^z + z)dz = ∫0^(3z-4) ∫0^2π (sinrcosθ - (3z-4))rdrdθ + ∫0^(3z-4) ∫0^2π ((rcosθ+cos(rcosθ))rdrdθ + ∫0^(3z-4) ∫0^2π ((e^z+z)rdrdθ

We can now integrate each term separately and evaluate the integral using the given limits. I would recommend using a calculator or computer software to help with the calculations.

For part (ii), we can use the divergence theorem to evaluate the surface integral. The divergence of the vector field F is given by divF = ∂/∂x (sinx-y) + ∂/∂y (x+cosx) + ∂/∂z (e^z+z). We can then use the divergence theorem to rewrite the surface integral as a volume integral: ∫∫s [(curlF) . nds] =