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I'm asked to calculate the work of the vector field $K(x,y,z)=(-y,x+z,-y+y^3+z)$ along the curve which is the intersection of the sphere of radius 1 and the plane $x=z$ :

a) directly

b) using Stoke's theorem

My problem is that I obtain different results in both cases.


For a) (correction : see edit 1 below), I use as parametrization $(cos(theta),sin(theta),cos(theta))$. The gradient is $(-sin(theta),cos(theta),-sin(theta))$. The dot product of that with our vector field in polar coordinates $(-sin(theta),2cos(theta),-sin(theta)+sin^3(theta)+cos(theta))$ is simply $2-sin^4(theta)-sin(theta)cos(theta)$. If we integrate that from $0$ to $2pi$, we get $frac{13pi}{4}$ noting that $sin(theta)cos(theta)$ vanishes.

For b), the curl of our vector field is $(3y^2-2,0,2)$. Our parametrization is $(rcos(theta),rsin(theta),rcos(theta))$. The cross product of the partial derivatives is simply $(-r,0,r)$. The dot product of that with our curl in polar coordinates is $4r-3r^3sin^2(theta)$. So now, we have $$int_{0}^{2pi}int_{0}^{1}(4r-3r^3sin^2(theta))rdrdtheta=piint_{0}^{1}(8r^2-3r^4)rdr=frac{31pi}{15}$$

So I don't see where I did something wrong, or where I forgot something.

Thanks for your help !

Edit 1:

Okay, as was pointed out in the comments and in one answer, I used a wrong parametrization in a). I should use $frac{cos(theta)}{sqrt{2}}$ instead of $cos(theta)$ for the parametrization of $x$ and $z$.

In this case, we get the dot product of $(-sin(theta), sqrt{2}cos(theta), -sin(theta)+sin^3(theta)+frac{cos(theta)}{sqrt{2}})$ and $(frac{-sin(theta)}{sqrt{2}}, cos(theta),frac{-sin(theta)}{sqrt{2}}) $ which yields $$frac{sin^2(theta)}{sqrt{2}}+sqrt{2}cos^2(theta)+frac{sin^2(theta)}{sqrt{2}}-frac{sin^4(theta)}{sqrt{2}}-frac{sin(theta)cos(theta)}{2}$$. If we integrate that from $0$ to $2pi$, the last term vanishes and using the common trig identity, we simply integrate $sqrt{2}-frac{sin^4(theta)}{sqrt{2}}$. So we get $frac{13pi}{4sqrt{2}}$. But it's still different fom b) though.

Edit 2 :

As was pointed out in the comments, I also need to use the new parametrization for part b). But even so, I get $frac{31pi}{15sqrt{2}}$

The cross product of the derivatives is $(frac{-r}{sqrt{2}},0,frac{r}{sqrt{2}})$. We need to integrate from $0$ to $2pi$ $dtheta$ and from $0$ to $1$ $dr$ the following (the curl remains the same): $(frac{4r}{sqrt{2}}-frac{3r^3sin^2(theta)}{sqrt{2}})r=frac{4r^2}{sqrt{2}}-frac{3r^4sin^2(theta)}{sqrt{2}}$ which yields $frac{8pi r^3}{3sqrt{2}}-frac{3pi r^5}{5sqrt{2}}$ evaluated from $0$ to $1$ and so we get
$frac{31pi}{15sqrt{2}}$

Edit 3 : problem solved. I multiplied by an extra $r$ in b). I should not have $(frac{4r}{sqrt{2}}-frac{3r^3sin^2(theta)}{sqrt{2}})r$ but simply $(frac{4r}{sqrt{2}}-frac{3r^3sin^2(theta)}{sqrt{2}})$ See my answer below.

Hint

Your parametrization is wrong in a) since it should be $$left({cos thetaoversqrt 2},{sintheta},{cos thetaoversqrt 2}right)$$

Okay, with the help of the comments, I found the error.

The first error in a) was that I should have used $(frac{cos(theta)}{sqrt{2}},sin(theta),frac{cos(theta)}{sqrt{2}})$ as parametrization in order to satisfy both constraints, i.e. $x=z$ and $x^2+y^2+z^2=1$. My parametrization only satisfied $x=z$.
With that, we get the correct answer, which is $frac{13pi}{4sqrt{2}}$

In b), my error was the same. I should also have used this parametrization in order to satisfy both constraints.

My second error in b) was to multiply the resulting dot product by an extra $r$. I thought this $r$ is the area element like in $rdrdtheta$. But if I take the cross product of the partial derivatives of the parametrization, this $r$ is already "included" in the entire thing (think of the determinant of the Jacobian) , so I don't need to multiply the entire thing again by that $r$. So I shouldn't have $(frac{4r}{sqrt{2}}-frac{3r^3sin^2(theta)}{sqrt{2}})r$ but simply $(frac{4r}{sqrt{2}}-frac{3r^3sin^2(theta)}{sqrt{2}})$. With that, once integrated, I also get $frac{13pi}{4sqrt{2}}$.

I hope that this will help someone in the future facing a similar problem.