# Surface area (cylindrical coordinates)

I'm having troubles when calculating the area of part of surface bonded with:$$x^2+y^2+z^2=a^2, x^2+y^2=ax, x^2+y^2=-ax.$$ The other two cylinders should give me the bounds for integral, so after I set cylindrical coordinates $$x=r\cos\phi, y=r\sin\phi$$ for first cylinder I get $$-\frac{\pi}{2}\le \phi\le\frac{\pi}{2}$$ and for the other cylinder $$\frac{\pi}{2}\le \phi\le \pi$$. Now not sure what of those two to use when calculating the surface area.

It is not very clear which surface area the question seeks you to find but given it says bounded by all three of them, it is more likely than not that it is seeking surface area of the sphere bounded between both cylinders. If not, we subtract it from $$4 \pi a^2$$ and that should give us the sum of surface area bound between each cylinder and sphere.

As you would notice both cylinders are of radius $$\frac{a}{2}$$ with centers at $$(-\frac{a}{2}, 0, z)$$ and $$(\frac{a}{2}, 0, z)$$ and are tangent to each other on $$z-$$axis.

It is much easier to do this in spherical coordinates. The surface of the sphere is defined by

$$x = a \cos \theta \sin \phi$$, $$y = a \sin \theta \sin \phi$$, $$z = a \cos \phi$$ $$(0 \leq \phi \leq \pi, 0 \leq \theta \leq 2 \pi) \,$$ where $$\phi$$ is the polar angle and $$\theta$$ is azimuthal angle.

Now we know that both cylinders intersect the surface of the sphere such that,

At $$z = 0 \,(or \, \phi = \frac{\pi}{2}): x = a, y = 0$$ for one of the cylinders and and $$x = a, y = 0$$ for the other. Similarly at $$z = a \, (or \, \phi = 0)$$, they both intersect sphere at $$(0, 0, \pm a)$$

From equation of the first cylinder and sphere, $$\,x^2 + y^2 = ax \implies a^2 \sin^2 \phi = a^2 \cos \theta \sin \phi$$

At intersection $$\theta = \cos^{-1} (\sin \phi)$$. For the second cylinder, $$\theta = \cos^{-1} (-\sin \phi)$$

So if we consider the surface only bound between $$0 \leq \theta \leq \pi$$ and $$0 \leq \phi \leq \frac{\pi}{2}$$, we can write

$$\theta = \frac{\pi}{2} - \phi$$ at the intersection of sphere and the first cylinder.

Similarly, $$\theta = \frac{\pi}{2} + \phi$$ at the intersection of sphere and the second cylinder.

So surface area of the surface bounded between three of them is given by

$$S = 4a^2 \displaystyle \int_0^{\pi/2} \int_{\pi/2 - \phi}^{\pi/2 + \phi } \sin \phi \, d\theta \, d\phi = 8a^2$$

If you are doing it in cylindrical coordinates, equations of the surfaces are $$r^2 + z^2 = a^2, r = \pm a \cos \theta$$. Based on one of the previous questions I saw from you, you already know how to find surface area in cylindrical coordinates. So leaving those details aside, your integral will be,

$$4 \displaystyle \int_{0}^{\pi} \int_{a |\cos \theta|}^a \frac{a \, r}{\sqrt{a^2-r^2}} dr \, d\theta$$

• Okey, I get i. Just why we set angle to be from $0$ to $\pi$(shouldn't it be whole circle). I mean, i get we have one circle in first and the other in second quadrant, but for some values in $[0,\pi]$ $\cos\phi$ and $-\cos\phi$ would be negative right? And why are we multipyling area with $4$? – Milica Koprivica Jan 6 at 22:31
• Yes $0$ to $\pi$ finds surface area in the first and second quadrant (using $2$D projection) and we are multiplying by $2$ as your rightly said there is one in third and fourth quadrant as well. Next, this is surface area for the hemisphere $z = \sqrt{a^2-r^2}$, you have another hemisphere below, $z = - \sqrt{a^2-r^2}$ and so another multiplication by $2$ (that makes multiplication by $4$). – Math Lover Jan 7 at 6:18

well: $$x^2+y^2+z^2=a^2$$ is a sphere of centre $$(0,0,0)$$ and radius $$a$$ then: $$x^2+y^2=ax\Rightarrow (x-a/2)^2+y^2=a^2/4$$ so it is a cylinder where the cross- section is centre $$(a/2,0,z)$$ and the radius is $$a/2$$ $$x^2+y^2=-ax\Rightarrow (x+a/2)^2+y^2=a^2/4$$ so this is a cylinder where the cross-section is centre $$(-a/2,0,z)$$ and the radius is $$a/2$$.

It is useful to look at the first by looking down on the $$x-y$$ plane where we see 3 circles, where the big overlaps the others at $$z=0$$ and as we move towards $$z=a/2$$ they only touch at a single point.

We will have $$-\pi/2\le\theta\le\pi/2$$ since we want to go the whole way around the $$z$$-axis so now you need to find a nice expression for $$r$$ which will most likely be dependent on both $$z,\theta$$