# Where does the error lie here?

Initially assume that the area $$a$$ and circumference $$c$$ of a radius $$R>0$$ circle are given axiomatically as:

$$a=πR^2$$ and $$c=2πR$$

Now in order to calculate the surface area of the top hemisphere (without its base) of a radius $$R$$ sphere, we approximate it with the following sum:

$$A=\sum_{i=1}^n2\pi r_iR\Delta\theta _i$$

Where $$r_i$$ is the radius of the disk on the plane $$z=Rcos{\theta}_i$$ and $${\theta}_i$$ is the complementary to the polar angle of a point that is both on that plane and the hemisphere surface.

$$A'$$ as it can easily be proven, when our partition length $$\Delta\theta_i$$ tends to zero for every $$i$$, converges to the integral:

$$\int_{0}^{\frac{π}{2}}2πR^2cos\theta d\theta=2πR^2$$

The surface area of the entire sphere is naturally twice that, hence the familiar formula:

$$A=4πR^2$$

Now it would be tempting to approximate the volume of the hemisphere by partitioning it as indicated by the following sum:

$$V'=\sum_{i=1}^nπ{r_i}^2R\Delta{\theta}_i$$

We get the area of the disk at a given point and we multiply that by the arc length that is directly in-between that disk and the one directly after it, with respect to our partitioning. Basically, a direct continuation of the process we followed for the derivation of the surface area of the sphere, instead there we had the circumference of the disk, whereas now we substitute that for its area.

But if we get the Riemann integral that corresponds to the sum $$V'$$, we get an obviously false result for the volume of the hemisphere:

$$V=πR^3\frac{π^2}{4}$$

When we know that actually: $$V=πR^3\frac{2}{3}$$

(Again, the last two values are associated with the hemisphere, ie half the sphere)

So, it is evident that there is something wrong with the assumption that we can generalize the first process, to derive the volume of the hemisphere. My question in a few words is:

Where, exactly, does that assumption fail?

When you calculated the area, $$R\Delta\theta_i$$ was a segment along the surface, "perpendicular" to the circle of radius $$r_i$$. If you follow the same procedure, you would need to multiply the area of the disk with radius $$r_i$$ with a segment perpendicular to the disk. $$R\Delta\theta_i$$ is tilted at an angle $$\pi/2-\theta$$. To get the perpendicular component, you need to multiply by $$\sin\theta$$. Then $$V=\int_0^{\pi/2}\pi R^3\sin^2\theta\sin\theta d\theta=\pi R^3\frac 23$$
• Your observation is accurate and the amendment, yields expected result. It's the same as to say that we integrate along the $z$-axis. But that doesn't really answer my question. My problem is that the process is derived by a direct continuation of the 'area process', ie in the approximation sums, the only difference being that one has the circumference and the other the area of a disk. The $R\Delta {\theta}_i$ segments are not really any more perpendicular to the disk circumference, than they are to the disk area. They have the exact same relative orientation in both instances. Nov 14, 2020 at 23:42
• Not true. Let's simplify the problem. Take a rectangular cuboid, where the height is much larger than the sides. Say $a=b\gg c$. The area is $(2a+2b)c$, and the volume is $abc$. Now tilt the sides in, let's say at $45^\circ$. Now you get a frustum. The area is about the same as before, but the volume decreased by a factor of $\sqrt 2$. That is because the distance perpendicular to the $ab$ plane decreased, but the distance along the face is the same. Nov 15, 2020 at 0:04
• Continuation to my previous comment: Why were you allowed to write in the first sum the $(2\pi r_i)(R\Delta\theta_i)$ as the area? Because in this context the two lines, one tangent to the circle in the plane of the circle, and the direction along $R\Delta\theta_i$ are perpendicular. Otherwise the formula for the area is incorrect. Nov 15, 2020 at 3:48