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Find the area of the cylinder given by $x^2+(y-1)^2=1$ between the xy-plane and the cone $\sqrt{x^2+y^2}+z=0$.

So parameterized the curve that describes the cylinder in the xy-plane:

$r(t)=(cost)i+(sint+1)j$

and then the height of surface must be $z=\sqrt{x^2+y^2}=\sqrt{2+2sint}$

I then solve the integral $ \int_{0}^{2\pi} \sqrt{2+2sint} \,dt$

But this does not give me the correct answer, which should be 16.

Edit (solution from my book that I don't understand):

The area asked for can be found by $$\begin{equation} \iint_{S_1}\,dxdy = 2\int_{C}1\,dS \end{equation}=2\int_{C}z\,ds$$ where $S_1$ is the surface of the cylinder between the xy-plane and the cone. $C$ is the curve of the cylinder in the first quadrant in the xy-plane given by $r(\theta)=2sin\theta$ for $0\leq \theta \leq \pi$ and dS is the areaelement across the curve with height z and width ds.

Then $ds=\sqrt{r(\theta)^2+(\frac{dr(\theta)}{d\theta})^2}=2d\theta$

So

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Mathomat55
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    Maybe I'm missing something, but the maximum height of your shape should be $z=2$, which means your area should be upper-bounded by $(2)2\pi\approx12.4$. Perhaps you're meant to find the entire area of the solid bounded by the 3 surfaces given, rather than the area of the cylidrical surface bounded by the others? – Angelica Nov 25 '21 at 14:18
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    Your answer of $8$ is absolutely correct – Math Lover Nov 25 '21 at 14:41
  • Thank you both for answering. It's weird that the solution says it should be 16. I can edit my original post and add the solution from the book. I don't understand the solution that well by the way. – Mathomat55 Nov 25 '21 at 16:30
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    I think I understand the books solution, and the hangup comes from how $C$ is defined. They say that $C$ is supposed to be the circular arc which is the projection of the cylinder onto the first quadrant of the xy-plane. However, it parametrizes this curve for $0\leq t\leq\pi$, which corresponds to the entire top half of the plane. As a result, instead of doubling half the area to get the final area, the book ends up just doubling the entire area. – Angelica Nov 25 '21 at 17:37
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    I could be mistaken however, so I'd check with someone else (a professor, perhaps) before asserting that your book is in error. – Angelica Nov 25 '21 at 17:39
  • @Angelica the book is in error and your analysis of mistake in the book is spot on. – Math Lover Nov 25 '21 at 19:50

1 Answers1

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The surface area you have found is absolutely correct. The book solution has an error. It defines curve $C$ as the curve of the cylinder in the first quadrant in the xy-plane. Now as the cylinder is in both first and second quadrants and the region is symmetric wrt y axis in xy-plane, it multiplies the answer by $2$. So it says -

$ \displaystyle \begin{equation} \iint_{S_1}\,dxdy = 2\int_{C}1\,dS \end{equation}=2\int_{C} |z| \,ds$

But if $C$ is only in the first quadrant in xy-plane, $0 \leq \theta \leq \frac{\pi}{2}$ and NOT $0 \leq \theta \leq \pi$.

Here are some more details that may explain to you the book solution. Your parametrization translates (no rotation) the coordinate axes such that to align the origin to the point $(0, 1)$ in the original coordinate system. The book uses the existing coordinates in which the cylinder is only in the first and second quadrants as seen in xy plane, and the parametrization uses polar coordinates $x = r \cos\theta, y = r\sin\theta, 0 \leq \theta \leq \pi$.

The equation of cylinder can be rewritten as $x^2 + y^2 = 2y \implies r = 2 \sin \theta, 0 \leq \theta \leq \pi$ in polar coordinates.

The equation of the cone is $z = - \sqrt{x^2+y^2} = - r$.

Now using the arc length formula in polar coordinates,

$ \displaystyle ds = \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} ~ d\theta$

As $r = 2 \sin\theta, ds = 2 ~ d\theta$

So the integral to find surface area $ \displaystyle \int_C |z| ~ ds$ can be rewritten as

$ \displaystyle \int_0^{\pi} 4 \sin\theta ~ d\theta = 8$

Or, $ ~ \displaystyle \int_0^{\pi/2} 8 \sin\theta ~ d\theta = 8$

Math Lover
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    Thank you so much for the help! I thought for a moment that I didn't understand the problem good enough because of the solution in the book. Thank you again :)¨ – Mathomat55 Nov 26 '21 at 16:29