Not Your Usual Cylinder - Surface Area

To calculate the surface area of an oblique cylinder, it is tempting to use a formula similar to the one used for a right circular cylinder, such as $2\pi r (r + l )$ or $2 \pi r (r + h )$ . However for reasons I shall explain ahead, it would not be correct to do so.

The oblique cylinder consists of three surfaces: Two flat surfaces and one curved surface. Each flat surface is a circle with radius $r$ , so the total area of flat surfaces is $2 \times \pi r ^{2}$ .

Now we will find the area of the curved surface. Let us first define a few terms.

• The axis of the cylinder is the line joining the centers of the flat surfaces.
• A perpendicular cross section of the cylinder is a cross section that is perpendicular to the axis of the cylinder.

Lemma : The perpendicular cross sections of the oblique cylinder are elliptical and not circular.

Proof : Choose the coordinate axes such that the axis of the cylinder coincides with a coordinate axis, say the z-axis. The curved surface is now parallel to the z-axis, and the perpendicular cross sections are parallel to the xy plane.

In the following figure, we see the flat surface is highlighted in blue, and a perpendicular cross section is highlighted in green.

We know that the flat surface is a circle, and the perpendicular cross section is a projection of the circle, therefore the perpendicular cross section is an ellipse and not a circle. $_\blacksquare$

Next we will cut a part of the cylinder along a perpendicular cross section and translate it as shown in the following figure. Observe that neither the lateral surface area, nor the volume of the cylinder is affected by this transformation. However, we obtain a right elliptical cylinder with the same perpendicular height as the slant height of our original oblique cylinder!

The area of the curved surface can now be found using "Perimeter of elliptical cross section × Height", i.e. $P \times l$ , where $P$ is the perimeter of the elliptical cross section. $P$ depends on the lengths of the semi-major and the semi-minor axis of the ellipse, so we will calculate those now. Let $a$ and $b$ be the semi-major and the semi-minor axes of the ellipse respectively.

Since we had projected the blue circle onto the green ellipse, i.e. we had projected along the y axis onto the xy plane, distances parallel to y axis are unchanged after the projection, and distances along the x axis are reduced. Thus the semi-major axis of the elliptical section is parallel to the y axis and is equal to the radius of the flat surface, $r$ , and the semi-minor axis is parallel to the x axis. Now we will find the semi-minor axis of the ellipse.

We will now use the invariance of volume to calculate the length of the semi-minor axis of the ellipse. Recall that the area of an ellipse with semi-major axis and semi-minor axis $a$ and $b$ is $\pi a b$ .

Also recall that the volume of the original oblique cylinder could be written as "Area of base × Perpendicular height" $= \pi r^{2} \times h$ .

Similarly, the volume of the newly formed right elliptical cylinder is $\pi a b \times l$ . Since the volume of both the cylinders is the same, we can equate them.

$\pi r^{2} \times h = \pi a b \times l$

We have already derived that $a = r$ , so let's substitute it in the above equation and make $b$ the subject of the formula.

$\pi r^{2} \times h = \pi r b \times l \implies b = r \times \frac{h}{l} ~~~~~~~~~~ \color{#3D99F6}{(1)}$

We see that the semi-major axis of the ellipse is $r$ , and the semi-minor axis is $r \times \frac{h}{l}$ .

Now that we know $a$ and $b$ in terms of the given data, we can proceed to find the value of $P$ . Unfortunately a closed form of the perimeter of ellipse does not exist. We can make use of an infinite series, such as the Gauss-Kummer Series to get the approximate value of the perimeter. The first few terms of the series are

$P = \pi ( a + b ) \left( 1 + \frac{1}{4} k^{2} + \frac{1}{64} k^{4} + \frac{1 } {256 } k^{6} + \cdots \right) ~~~~~~~~~~~~~ \color{#3D99F6}{(2)}$

where $k = \dfrac{a -b } { a + b }$ .

After substituting values of $r, h$ and $l$ in equation $\color{#3D99F6}{(1)}$ , we get $a = 2$ and $b = \frac{8}{5}$ . Using this, we see that $k = \frac{ 1 } {9}$ .

Substituting values of $a, b$ and $k$ in equation $\color{#3D99F6}{(2)}$ , we obtain $P \approx 11.34467$ .

The total surface area of the oblique cylinder is the sum of the areas of the flat surfaces and the curved surface.

$S = 2 \times \pi r^{2} + P \times l ~~~~~~~~~~\color{#3D99F6}{(3)}$

After substituting $r, l$ and $P$ in equation $\color{#3D99F6}{(3)}$ , we get $S \approx \boxed{81.856 }$ $_\square$

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To conclude, the only correct way to find the lateral surface area of any cylinder whose flat surfaces are parallel and congruent to each other (irrespective of whether the cylinder is right or oblique) is

$\text{Lateral Surface Area} = \text{ Perimeter of Perpendicular Cross Section} \times \text{ Length of Axis }$

Using the perimeter of any other cross section will yield incorrect results. In case of the right cylinder, the flat surface is the same as the perpendicular cross section so it is not much of an issue there. However, special care must be taken in the case of an oblique cylinder, where the flat surfaces and the perpendicular cross sections are not the same.

The formulae stated at the beginning of the solution use a circular cross section. The circular cross section is not perpendicular to the axis of the cylinder, so it gives an incorrect answer. This concept is nicely summarized on this webpage: http://www.mathalino.com/reviewer/solid-mensuration/cylinder

Wonderful problem Pranshu! I solved it by a pretty different method however... i'll just mention the method w/o getting into calculations.

• The lateral surface of the cylinder can be broken down into a series of 'strips' as follows...
• Establish a polar coordinate system on the top and bottom circles such that the reference axes are parallel to each other. Now consider a small section of the circumference at $\theta$ of length $rd\theta$ for both the circles. Thus we have a 'strip' joining these two small segments of length $rd\theta$ .
• The length of this strip is $l$ (slant height) but width is not $rd\theta$ . This is where the confusion arises from.
• We need to find the angle $\phi$ between $rd\theta$ and $l$ as a function of $\theta$ . Then the required lateral area equals $\int _{ 0 }^{ 2\pi }{ l } \cdot sin(\phi (\theta ))\cdot rd\theta$ ... where $\phi (\theta )$ is the angle mentioned above.
• $\phi (\theta )$ can be calculated by many methods but I preferred using vectors. Adopting a rectangular coordinate system the direction vector for $rd\theta$ becomes $sin\theta i + cos\theta j$ and vector for $l$ becomes $3i + 4k$ . Thus by dot products, etc. we get $cos\phi = (3/5)sin\theta$ .
• Now its simply a job of putting everything together into that integral and evaluating it to get the answer and of course adding onto it the circular area part.

P.S. How to put those sketches into the text? Is there any software that you can use to make those sketches? (My solution would be better off with a diagram)