Volume resulting from revolving a rectangle around a vertical line

Question

As shown in the figure, I have a rectangle with sides a and b and I want to revolve it around the z-axis (vertical line) creating a solid cylinder.

The volume (V) of the cylinder is well known and is $$V=\pi r^2h$$

where r is the radius of the cylinder and h is the height of the cylinder. In this case, length a of the rectangle side is the radius and b is the height. But if I now rotate this rectangle around the vertical axis through the point in the center and try to integrate to get the volume, isn't the differential volume equal to,

dV = Area of Rectangle multiplied by the differential thickness = (ab)(Rd$\phi)$

where $\phi$ is the angle of rotation and $Rd\phi$ is the differential thickness of the arc over which the rectangle moves. But since R is equal to a, plugging in yields,

$$dV = abad\phi = a^2bd\phi$$

The volume is then found by integrating from $0$ to $2\pi$,

$$V = \int_{0}^{2\pi}a^2bd\phi$$

and since a and b are constant,

$$V = a^2b\int_{0}^{2\pi}d\phi$$

or

$$V = a^2b(2\pi - 0) = 2\pi a^2b$$

Now comparing this to the standard formula for a cylinder by substituting in r = a and h = b yields,

$$V = 2\pi r^2h$$

This result is twice the standard formula for a cylinder. Where did I go wrong?

Edit#1: I have read the answers on this question. But this was just a very simple geometry to help me understand. What I really want to do is rotate a variable area trapezoid around a focus of an ellipse and obtain an expression for the volume. Assume the side of the trapezoid through the point is constant and equal to "c". In the x-y plane that cuts the solid in half the distance is "a" (this would be equivalent to the height of a trapezoid when figuring the area) and b is the height along the edge of the ellipse. So, the area of the trapezoid would be,

$$Area = \frac{c+b(\phi)}{2}a(\phi)$$

and I thought the differential distance would be $a(\phi)d\phi$

$$dV = \frac{c+b(\phi)}{2}a(\phi)a(\phi)d\phi=\frac{c+b(\phi)}{2}(a(\phi))^2d\phi$$

I've included a .gif below which I hope is viewable.

Edit#2: Based on the responses for the rotating rectangle, I'm now thinking that the differential volume for the rotation of the variable area trapezoid should be,

$$dV = \frac{(a(\phi))^2b(\phi)}{2}d\phi$$

If someone could confirm this or tell me why this is incorrect I would be most grateful. Thank you.

Edit#3: As I think about this some more in light of the answers below, I think I still have a problem in that the height, "b" is not uniform over the length "a, where "a" now is any internal length of the vector (at a specific angle of $\phi$) from the point, through which the vertical axis runs, to the ellipse edge. So, "b" is then a function of both "a" and $\phi$? If it makes more sense "a" could be replaced by "r" and then r = a at the edge of the ellipse. So using "r", the volume would be

$$dV = \frac{(a(\phi))^2b(r,\phi)}{2}dr d\phi$$

and now I need a double integral. Any thoughts from anyone?

Answer 1

Think of the differential volume as the area of sector of angle $d\theta$ times the height $b$. It is well known that area of the sector is $dA=\frac{1}{2}a^2d\theta$. So the differential volume is just one half of what you thought. And therein lies your problem.

Answer 2

The edge labeled $b$ in your figure moves a distance $R\,\mathrm d\phi$ when the rectangle rotates by the angle $\mathrm d\phi,$ but the opposite edge doesn't move at all. So instead of a thin plate of uniform thickness you have a thin wedge, and it has half the volume that you assume it does.

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The factor of $\frac12$ works out in this case because you can take two identical wedges of this shape and paste them together to make a plate of uniform thickness. Alternatively you can consider the wedge to be a triangular prism; this gives the same result.

Neither of these methods works in general for wedges of other shapes. In particular, if the two large faces are trapezoids with unequal bases, the volume will not be $\frac12$ of the volume of a thin plate of the same maximum thickness. The volume will be greater than $\frac12$ the volume of the plate if the smaller base forms the "sharp" edge of the wedge, and less if the larger base forms the "sharp" edge.

Answer 3

"But since $R$ is equal to $a$" is not true.

When you sweep the rectangle, you have to consider the average radius, which is $\dfrac a2$. (The swept volume increases linearly with the distance to the axis.)

Answer 4

The basic idea in a general case is to consider a differential wedge whose height at the axis of rotation is some function $a(\theta)$ of a rotation angle $\theta \in [0,2\pi)$, whose radius is some function $r(\theta)$, and whose height at $r$ is $b(\theta)$. The angle measure at its apex is $d\theta$.

To calculate the differential volume of this wedge, we slice it into rectangles of differential width $dx$. Their height is a function $f(x)$ of the radial distance $x \in [0, r(\theta)]$. In your case, $$f(x) = \left(1 - \frac{x}{r(\theta)}\right)a(\theta) + \frac{x}{r(\theta)} b(\theta); \tag{1}$$ i.e., it is a linear interpolation of the endpoints. The width of a representative rectangle as a function of the radial distance is simply $x d\theta$ (when $\theta$ is measured in radians). Consequently, the differential volume of such a wedge is $$dV(\theta) = \int_{x=0}^{r(\theta)} f(x) (x d\theta) \, dx = \frac{r^2(\theta)}{6}(a(\theta) + 2b(\theta)) \, d\theta. \tag{2}$$ All that remains is to integrate this over $\theta \in [0,2\pi)$ to obtain the total volume: $$V = \frac{1}{6} \int_{\theta = 0}^{2\pi} r^2 (\theta) (a(\theta) + 2b(\theta)) \, d\theta. \tag{3}$$ We can verify in the case $a = b = \text{constant}$ and $r = \text{constant}$, we recover the volume of a cylinder with radius $r$ and height $a$.

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In your specific case, $a$ is a constant function of $\theta$, representing the height of the solid at the axis of rotation, but $r$ and $b$ are variable functions whose forms depend on your application. In the case where the axis of rotation is situated at a focus of an ellipse with semimajor axis $p$ and semiminor axis $q$, i.e. the boundary is parametrized by $$(c + p \cos t, q \sin t), \quad c = \sqrt{p^2 - q^2}, \tag{4}$$ we must be careful to not assume $\theta = t$. The reason is that $t$ parametrizes the angle in a transformed coordinate space where the ellipse maps to a circle; thus, when $p \ne q$, the angle of integration $\theta$ will not correspond. Thus we require a reparametrization: the line $y \cos \theta = x \sin \theta$ intersects the ellipse for a value of $t$ satisfying $$(c + p \cos t)\sin \theta = q \sin t \cos \theta, \tag{5}$$ or $$t = \tan^{-1} \frac{q(p + c \cos \theta) \sin \theta}{q^2 \cos \theta - cp \sin^2 \theta}, \tag{6}$$ hence the ellipse is reparametrized as a function of the angle relative to the focus as: $$(x(\theta), y(\theta)) = \frac{2q^2 (p + c \cos \theta)}{p^2 + q^2 - c \cos 2\theta} (\cos \theta, \sin \theta). \tag{7}$$ While this parametrizes the boundary of the region to be integrated, we only need its distance from the axis of revolution; i.e., $$r(\theta) = \sqrt{x^2(\theta) + y^2(\theta)} = \frac{2q^2 (p + c \cos \theta)}{p^2 + q^2 - c \cos 2\theta}. \tag{8}$$

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Next, we need a boundary height function $b(\theta)$. If the solid is to have planar surfaces, it needs to have a height function that depends only on the $x$-coordinate of the parametrized ellipse; e.g., $$b(\theta) = a + (b^* - a) x(\theta). \tag{9}$$ This choice of $b$ gives $b(0) = b^*$ and $b(\pi/2) = a$ where $b^*$ is an extremal height. Then the other extremum of height occurs when $\theta = \pi$, giving $b(\pi) = a + (b^* - a)(c - p)$. All that remains is to set up the integral, remembering that we want twice $(3)$ due to symmetry about the plane of integration:

$$2V = \frac{1}{3} \int_{\theta = 0}^{2\pi} \left(\frac{2q^2 (p + c \cos \theta)}{p^2 + q^2 - c \cos 2\theta}\right)^2 \left(3a + 2(b^*-a)\frac{2q^2 (p + c \cos \theta)}{p^2 + q^2 - c \cos 2\theta} \cos \theta \right) \, d\theta. \tag{10}$$

Applying this formula in practice for the case $a = 3$, $b^* = 4$, $p = 2$, $q = 1$ yields $2V = 4(3 + \sqrt{3}) \pi$.

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A final remark about this question. For the elliptical region with planar ends, I feel like a superior approach would be to compute the volume as a suitably transformed circular cylinder. Note I have not done such a computation, nor have I checked the above for accuracy.