Surface area of sphere $x^2 + y^2 + z^2 = a^2$ cut by cylinder $x^2 + y^2 = ay$, $a>0$

Question

The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$.

The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates.

We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have:

$$A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA$$

where $F$ is the equation of the sphere.

Plugging in the expressions and simplifying ($z \ge 0)$, we get:

$$A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy$$

Converting to polar coordinates, we have:

$$A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta$$

Calculating this I get $2\pi a^2$. The answer is $(2\pi - 4)a^2$. Where am I going wrong?

Answer 1

Given the equations

$$ x^2+y^2+z^2=a^2, $$

and

$$ x^2+y^2 = ay, $$

we obtain

$$ ay + z^2 = a^2. $$

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Using

$$ \begin{eqnarray} x &=& a \sin(\theta) \cos(\phi),\\ y &=& a \sin(\theta) \sin(\phi),\\ z &=& a \cos(\theta),\\ \end{eqnarray} $$

we obtain

$$ a^2 \sin(\theta) \sin(\phi) + a^2 \cos^2(\theta) = a^2 \Rightarrow \sin(\theta) = \sin(\phi) \Rightarrow \theta=\phi \vee \theta=\pi-\phi. $$

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For the surface we have

$$ \begin{eqnarray} \int d\phi \int d\theta \sin(\theta) &=& \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta) + \int_{\phi/2}^{\pi}d\phi \int_0^{\pi-\phi} d\theta \sin(\theta)\\ &=& 2 \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta). \end{eqnarray} $$

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We can calculate the surface as

$$ \begin{eqnarray} 4 a^2 \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta) &=& 4 a^2 \int_0^{\pi/2}d\phi \Big( 1 - \cos(\phi) \Big)\\ &=& 4 a^2 \Big( \pi/2 - 1 \Big)\\ &=& a^2 \Big( 2\pi - 4 \Big). \end{eqnarray} $$

Answer 2

I know what you r doing rong, I solved this a week ago, the same way you did.

You forgot that $\sqrt{\sin^2\theta} = |\sin\theta|$, not $\sin\theta$

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This is how you might have done

$$ 2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos\theta}\dfrac{a}{\sqrt{a^2-r^2}}\cdot r \cdot drd\theta$$

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$$= 2\int_{-\pi/2}^{\pi/2}\left[ -a\sqrt{a^2-r^2}\right]_0^{a\cos\theta}d\theta$$

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$$= 2\int_{-\pi/2}^{\pi/2}- a^2\sqrt{\sin^2\theta}-\left(-a^2 \right)d\theta$$

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$$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2\sin\theta d\theta$$

MISTAKE !!!!

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INSTEAD $$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2|\sin\theta| d\theta$$

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$$= 4\int_{0}^{\pi/2}a^2- a^2\sin\theta d\theta$$

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$$= 4a^2 \left[1- \sin\theta \right] _{0}^{\pi/2}$$

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$$= a^2\left(2\pi-4 \right)$$

Answer 3

The integral should be from 0 to and not -/2 to /2 as shown in problems above. So your integrand is correct with right limits. Now, take |cos| in last step so you get integral cos from 0 to /2 and integral -cos from /2 to .