# Closed line integral with spherical coordinates

So I had this question in my homework, but I could not solve it.

The question is as follows:

There is a closed path going from (0,0,0) to (1,0,0) to (0,1,0) to (0,1,2) to (0,0,0). I have first transformed this to spherical coordinates This gets me (with $$(r,\theta,\phi)$$ with theta the angle with the z-axiom and $$\phi$$ with the x): $$1:(0,0,0)$$$$2:(1,\pi/2,0)$$$$3:(\sqrt5,\arctan(1/2),\pi/2)$$$$4:(0,0,0)$$ There is a vector field : $$\mathbf A = r\cos^2\theta\,\boldsymbol{\hat r}-r\cos\theta\sin\theta\,\boldsymbol{\hat\theta}+3r\,\boldsymbol{\hat\phi}$$ And I have to solve the line integral of A over this path. I have tried solving it by dividing the closed integral in 4 different ones and summing those up. For each of these integrals I have defined a different ds-vector. For the first one , $$ds=rd\phi\boldsymbol{\hat\phi}$$ For the second one, $$ds=dr\hat r$$ For the 3rd and 4th ones, $$ds=rd\theta\hat\theta+dr\hat r$$

I struggle with the 3rd and 4th integrals, because I only have a d$$\theta$$ but in my integral r isn't constant in this interval; I don't know how to solve this. I thought of entering my r values together with my $$\theta$$ values even though I don't integrate those. Doing this I came up with the answer $$1+1.5\pi$$ , but the correct answer is only $$1.5\pi$$ Could someone help me out?

Hi I'm really sorry I forgot to mention that all paths are straight lines, except the one from (1,0,0) to (0,1,0). This one is a quarter of a circle.

• $1.5 \pi$ is not correct either. Are all the paths along straight lines? Have you typed the path and the vector field correctly? Apr 7, 2021 at 5:37
• Hi I'm really sorry I forgot to mention that all paths are straight lines, except the one from (1,0,0) to (0,1,0). This one is a quarter of a circle. Apr 7, 2021 at 7:51

Vector field $$\vec F = r\cos^2\theta\,\boldsymbol{\hat r}-r\cos\theta\sin\theta\,\boldsymbol{\hat\theta}+3r\,\boldsymbol{\hat\phi}$$

Th given path $$\gamma: A(0, 0, 0) \to B (1, 0, 0) \to C(0, 1, 0) \to D (0, 1, 2) \to A \$$ (where all are straight lines except path $$B \to C$$ which is on a circle).

An easier solution is to convert this into two closed paths and do double integral,

$$\gamma_1: A \to B \to C \to A$$ in $$XY$$ plane ($$BC$$ is a circular arc)
$$\gamma_2: A \to C \to D \to A$$ in $$YZ$$ plane

Please note segment $$C \to A$$ in path $$\gamma_1$$ and $$A \to C$$ in path $$\gamma_2$$ and line integrals over these segments cancel out each other and hence $$I(\gamma) = I(\gamma_1) \cup I(\gamma_2)$$.

Now using Stokes' Theorem,

$$\displaystyle I(\gamma_1) = \int_{\gamma_1} \mathbf A \cdot dr = \int_{ABC} (\nabla \times \vec F) \cdot \hat n \ dS \$$

$$\displaystyle I(\gamma_2) = \int_{\gamma_2} \mathbf A \cdot dr = \int_{\triangle ACD} (\nabla \times \vec F) \cdot \hat n \ dS$$

Using the formula for curl in spherical coordinates, $$\nabla \times \vec F = 3 \cot \theta \ \hat r - 6 \ \hat \theta$$

As surface $$ABC$$ is in $$XY$$ plane, $$\theta = \frac{\pi}{2}$$ and $$\hat n = (\cos\theta \ \hat r - \sin\theta \ \hat \theta) = - \ \hat \theta$$

So the first integral is $$6$$ times area of surface $$ABC$$. Surface area of $$ABC$$ is quarter of a unit circle = $$6 \cdot \frac{\pi}{4} = \frac{3 \pi}{2}$$.

As surface $$ACD$$ is in $$YZ$$ plane, $$\phi = \frac{\pi}{2}$$ and,
$$\hat n = \sin\theta \cos\phi \ \hat r + \cos\theta \cos\phi \ \hat \theta - \sin \phi \ \hat \phi = - \hat \phi$$

So the second integral is zero.

That leads to the final answer of $$\frac{3 \pi}{2}$$.