I am learning vector calculus using Vector Calculus by Marsden and Tromba. In the chapter Integrals over Paths and Surfaces, I went across the following problem:
A restaurant is being built on the side of a mountain. The mountain's surface is given by the equation $z+x^2+y^2=4R^2$ and the hotel is the cylinder $ (y-R)^2+x^2=R^2 $. What will be the curved surface area of the hotel?
The book provides the following figure:
Originally the question has asked a lot more, but right now, I only have problems in this part.
Initially I had no idea how to approach the problem, but slowly I came up with this solution.
On the $yz$ plane, the surface of the hotel projects the region $R$ bounded by $z=4R^2$, $y=2R$, and $z=4R^2-y^2$. The surface of the hotel will be symmetric around the $yz$ plane. So, if we calculate the surface area of one of the sides, the whole surface area can be calculated easily. Along the positive $x$ axis, the surface of the hotel will be given as $x=\sqrt{R^2-(y-R)^2}$.
So, surface area of that side can be given as: $$ \begin{align} A(S) &= \iint_R \sqrt{ \Big(\frac{\partial x}{\partial y}\Big)^2+\Big(\frac{\partial x}{\partial z}\Big)^2+1}\,\mathrm dA \\ &= \int\limits_0^{2R}\int\limits^{4R^2}_{4R^2-y^2}\sqrt{\frac{(y-R)^2}{R^2-(y-R)^2}+1}\,\mathrm dz\,\mathrm dy\\ &= R \int\limits_0^{2R}\frac{y^2}{\sqrt{R^2-(y-R)^2}}\,\mathrm dy \\ & \overset{y-R=R \sin u}{=} R^3 \int\limits_{-\pi/2}^{\pi/2} (1+\sin u)^2 \,\mathrm du \\ A(S)&= \frac{3\pi R^3}2 \end{align}$$
Is this correct? If no, what is the mistake? If yes, are there any other ways to approach the problem? I don't think that it is correct, primarily due to the reason that it is not dimensionally correct. Except that, I have no good reason to say that the answer is incorrect. Maybe this is due to $z$ not having dimension of length, but that of area.
I tried to find the solution manual for the book, but didn't find any. Any help regarding that too would be appreciated.
It is simpler to see that the surface area element of a cylinder of radius $R$ is $dS = R \cdot d\theta \cdot dz$
The hotel surface is given by $(y-R)^2+x^2=R^2$. So it is easier to use the parametrization for cylindrical surface $x = R \cos\theta, y = R + R\sin\theta, z = z$
At the intersection of paraboloid surface and the cylinder,
$z = 4R^2 - x^2 - y^2 = 4R^2 - R^2 \cos^2\theta - R^2 (1 + \sin^2\theta + 2 \sin\theta) = 2R^2(1 - \sin\theta)$
So the curved surface area of the hotel is given by,
$ \displaystyle \int_0^{2\pi} \int_{2R^2(1 - \sin\theta)}^{4R^2} R ~ dz ~ d\theta = 4 \pi R^3$