This question is about line integrals in cartesian and cylindrical coordinates.
It is based on the following problem from Apostol's Calculus, Volume II, chapter 10 "Line Integrals", section 10.9
- Calculate the work done by the force field $$f(x,y,z)=y^2\hat{i}+z^2\hat{j}+x^2\hat{k}\tag{1}$$ along the curve of intersection of the sphere $x^2+y^2+z^2=a^2$ and the cylinder $$x^2+y^2=ax\tag{2}$$ where $$z\geq 0\tag{3}$$ and $$a>0\tag{4}$$ The path is traversed in a direction that appears clockwise when viewed from high above the $xy$-plane.
The answer from Apostol's book is $\pi a^3/4$.
Initial Attempt
At first I did everything in cartesian coordinates.
From (2) we have
$$y=\pm\sqrt{ax-x^2}\tag{5}$$
Subbing (2) into (1) and solving for $z$ we have
$$z=\sqrt{a^2-ax}\tag{6}$$
From (5) and (6) we have the condition $x\in [0,a]$.
If we let $x$ be our parameter, a parametrization of the intersection of the sphere and the cylinder is given by the two equations in (7) (notice the $\pm$ on the $\hat{j}$ component)
$$\vec{r}(t)=t\hat{i}\pm\sqrt{at-t^2}\hat{j}+\sqrt{a^2-at}\hat{k}\tag{7}, t\in [0,a]$$
Here is a plot of the two curves that give the full intersection (for $a=2$)
Hence
$$d\vec{r}=(\hat{i}\pm\frac{a-2t}{2\sqrt{at-t^2}}\hat{j}+\frac{-a}{2\sqrt{a^2-at}}\hat{k})dt\tag{8}$$
and
$$f(\vec{r}(t))=(at-t^2)\hat{i}+(a^2-at)\hat{j}+t^2\hat{k}\tag{9}$$
And the work done by the force field on this curve is
$$\int_C \vec{f}\cdot d\vec{r}=\int_0^a \vec{f}(\vec{r}(t))\cdot d\vec{r}(t)dt\tag{10}$$
$$=\int_0^a \left(at-t^2+\frac{(a^2-at)(a-2t)}{2\sqrt{at-t^2}}-\frac{at^2}{2\sqrt{a^2-at}}\right )dt \tag{11}$$
My first question is if this approach is correct so far. This integral does not seem easy, and using Maple to solve it doesn't yield an answer.
Second Attempt
Then I tried with cylindrical coordinates, a topic that still eludes me.
The sphere is
$$r^2+z^2=a^2\tag{12}$$
and the cylinder is
$$r^2=ar\cos{\theta} \implies r=a\cos{\theta}\tag{13}$$
From these two equations we obtain
$$z=\sqrt{a^2-r^2}=\sqrt{a^2-a^2\cos^2{\theta}}=a\sin{\theta}\tag{14}$$
So at this point we have a parametrization of the intersection.
$$r=a\cos{\theta}$$ $$\theta=\theta$$ $$z=a\sin{\theta}$$
And if we plot this, we get something that seems to make sense as well
except that we have to remember that our intersection has $z\geq 0$, so it our intersection is the top part of the plot above, and so coincides with the plot we obtained when we used cartesian coordinates.
Now is the part that is more murky for me.
We have
$$\vec{f}(\vec{r}(\theta))=a^2\cos^2{\theta}\sin^2{\theta}\hat{i}+a^2\sin^2{\theta}\hat{j}+a^2\cos^4{\theta}\hat{k}\tag{15}$$
$$\vec{r}(\theta)=r(\theta)\hat{r}(\theta)+z(\theta)\hat{k}\tag{16}$$
$$=a\cos{\theta}\hat{r}(\theta)+a\sin{\theta}\hat{k}\tag{17}$$
$$=a\cos^2{\theta}\hat{i}+a\cos{\theta}\sin{\theta}\hat{j}+a\sin{\theta}\hat{k}\tag{18}$$
Then
$$\frac{d\vec{r}(\theta)}{d\theta}=-2a\cos{\theta}\sin{\theta}\hat{i}+(-a\sin^2{\theta}+a^2\cos^2{\theta})\hat{j}+a\cos{\theta}\hat{k}\tag{18}$$
Thus, if this is correct, then the line integral is
$$\int_C \vec{f}\cdot d\vec{r}=\int_{\pi}^0\left ( -2a^3\sin^3{\theta}\cos^3{\theta}-a^3\sin^4{\theta}+a^4\sin^2{\theta}\cos^2{\theta}+a^3\cos^5{\theta} \right )d\theta\tag{19}$$
Note that I chose the integration limits by looking at the parameterized equations and figuring out the values of $\theta$ that trace out the trajectory we want.
For example, consider $a=2$, as in the plots above. If we start at $\theta=0$ which corresponds to $(r,\theta,z)=(2,0,0)$ and move to $\theta=\pi/2$ which is $(r,\theta,z)=(0,\pi/2,2)$. This is halfway. Then we go to $\theta=\pi$ which takes us back to the starting point, but now with coordinates $(-2,\pi,0)$. However, this is counterclockwise, and we want clockwise, as per the problem statement. Hence the limits go from $\pi$ to $0$.
Using Maple, (19) evaluates to
$$\frac{\pi a^3}{4}\tag{2}$$
So my questions are the following
- is the approach in Cartesian coordinates just destined to fail because that integral is too difficult to solve?
- is there some easier way than what I did above with the cylindrical coordinates?
Final Note: As sometimes happens, in the course of carefully writing out my question and making sure it is well organized and clear for the reader, I sometimes end up learning. And it seems that I did reach the correct result after all with the cylindrical coordinates.
However, my questions remain about if this was all way too cumbersome and if there is a smarter way.