Use Stokes' theorem to show that the line integral has the value given.

Question

Problem 9 from Apostol's Calculus 2 book, page 443:

Use Stokes' theorem to show that the line integral has the value given:

$$\int_C (y^2 + z^2)\,dx + (x^2 + z^2)\,dy + (x^2 + y^2)\,dz = 2\pi ab^2$$

Where $C$ is the intersection of the hemisphere $x^2+y^2+z^2 = 2ax$, $z>0$, and the cylinder $x^2 + y^2 = 2bx $, where $0

Attempt

As suggested by Ted Shifrin, I used the portion of the cylinder between the $xy$-plane and the intersection with the sphere.

In the intersection, we have: $$x^2+y^2 = 2bx$$ $$x^2 + y^2 + z^2 = 2ax $$

Thus, $$z^2 = 2(a-b)x$$

Since the curve of intersection is on the cylinder, I used the following parametrization: $$x = b + b\cos\theta$$ $$y= b\sin\theta $$ $$z=z$$

With $0 < z < \sqrt{2(a-b)b(1+\cos\theta)}$, because the surface is between the $xy$-plane and the intersection of the cylinder with the sphere. Thus,

$$\vec r(\theta, z) = (b+b\sin\theta, b\cos\theta, z) $$

Using the Stokes theorem, we have:

$$\int_C (y^2 + z^2)\,dx + (x^2 + z^2)\,dy + (x^2 + y^2)\,dz = \iint_S 2(y-z,z-x,x-y) \cdot \frac{\partial\vec r}{\partial \theta} \times \frac{\partial\vec r}{\partial z} \,dz \,d\theta$$

Where $S$ is the portion of the cylinder between the $xy$-plane and the intersection with the sphere. Besides, $$\frac{\partial\vec r}{\partial \theta} \times \frac{\partial\vec r}{\partial z} = b(\cos\theta, \sin\theta,0) $$

Therefore, using the parametrization, we have:

$$2b\iint_S (b\sin\theta - z, z - b - b\cos\theta, b + b\cos\theta - b\sin\theta)\cdot (\cos\theta, \sin\theta,0)\,dz \,d\theta = $$ $$2b\int_0^{2\pi}\int_0^{\sqrt{2(a-b)b(1+\cos\theta)}} [z(\sin\theta - \cos\theta) - b\sin\theta ]\,dz\,d\theta = -2\pi b^2(a-b)$$

Whose module is different from $2\pi ab^2$. What's wrong?

Answer 1

HINT: (slightly edited) You might try that portion of the cylinder lying between the $xy$-plane and the sphere, oriented correctly. The curve $C$ and the curve $C' = \{x^2+y^2=2bx, z=0\}$ (each properly oriented) will form the boundary of this surface, and so, keeping track of orientations, one will need to adjust by the line integral over $C'$. (Did Apostol specify the orientation of $C$? I bet yes.)