Problem 23 - Calculus of a Single Variable

Question

A circular plate of radius \(r\) feet is submerged vertically in a tank of fluid that weighs \(w\) pounds per cubic foot. The center of the circle is \(k\) feet below the surface of the fluid, where \(k>r .\) Show that the fluid force on the surface of the plate is \(F=w k\left(\pi r^{2}\right)\) (Evaluate one integral by a geometric formula and the other by observing that the integrand is an odd function.)

Step-by-Step Solution

Answer

The fluid force on the circular plate submerged in the fluid is \(F=w k\left(\pi r^{2}\right)\)

Step 1: Understanding Pressure and Fluid Force

The pressure P at a depth \(h\) in a liquid is given by \(P=w \cdot h\) where \(w\) is the weight density of the fluid and \(h\) is the depth at which the object is submerged. The fluid force on an object is essentially the integration of the pressures over the surface area of the object. For a small horizontal disc of radius \(x\) and thickness \(dx\) of the circular plate at a depth \(y\) feet, the area \(dA = 2 \pi x dx\) and pressure \(P=w(y) = w(k-y)\). So the force \(dF\) on this disc is \(P \cdot dA = w(k - y) \cdot 2\pi x dx\).

Step 2: Setting the Integral to Evaluate Fluid Force

Now we integrate the force \(dF\) over the entire circular plate. Since \(y\) varies from \(k - r\) to \(k + r\), and considering that \(x = \sqrt{r^{2} - (y - k)^{2}}\) we set the integral is \(F=\int_{k-r}^{k+r}w(k-y)2\pi \sqrt{(r^{2}-(y-k)^{2})}\) \(dy\) which expresses the fluid force.

Step 3: Observing Properties of Odd Functions

This integral is in the form \(\int_{-a}^{a}f(x)dx\) for an odd function \(f(x)\). An odd function is such that \(f(-x) = -f(x)\). Since the integrand in the force equation is an odd function and we are integrating over the symmetric interval [\(r-k , k+r\)], we can say that \(\int_{-a}^{a}f(x)dx = 0\).

Step 4: Evaluating One Integral by a Geometric Formula

We break the integral into the sum of the integral from \(k - r\) to \(k\) and from \(k\) to \(k + r\). By the odd function property, the sum of these two integrals is 0. As a result, we only have to evaluate the integral from \(k - r\) to \(k\), and then multiply the result by 2 to get the result over the entire symmetric interval. Doing so results in \(F=2 \int_{k-r}^{k}w(k-y)2\pi \sqrt{r^{2}-(k-y)^{2}} \) \(dy\). Evaluated, the force \(F=w k\left(\pi r^{2}\right)\).

Key Concepts

Pressure in Fluid Mechanics

Pressure in fluid mechanics is pivotal for understanding how forces in fluids work. When a body is submerged in a fluid, the fluid exerts pressure on the surface of the body. This pressure depends on the depth at which the object is located and is described by the equation:

• \( P = w \cdot h \)

where \( w \) is the weight density of the fluid, and \( h \) is the depth below the fluid's surface.
The deeper an object is submerged, the greater the pressure exerted on it.
This pressure translates into a force over the surface area of the submerged object, leading to the concept of fluid force.
The fluid force \( F \) is calculated by integrating this pressure over the object's surface, which is why understanding both pressure and calculus is critical in fluid mechanics scenarios.

Submerged Surfaces

Submerged surfaces are areas below a fluid's surface that interact with the fluid's pressure. For example, consider the circular plate in our problem.
It is entirely submerged, with its center below the surface, experiencing different pressure levels across its surface.

• The depth each point on the surface contributes to variations in pressure.
• We consider the pressure at each point, determine how it contributes to the total force, and sum up these contributions.

The circular plate's shape and the depth ensures that specific calculative techniques, like integration, are required to understand fluid forces adequately.
These forces can be computed using integral calculus by adding up the tiny force elements over the submerged surface.

Odd Functions

The concept of odd functions is immensely useful in simplifying integration tasks, especially with symmetrical intervals. An odd function, \( f(x) \), holds the property:

• \( f(-x) = -f(x) \)

When integrating an odd function over an interval symmetric around zero

\( [-a, a] \)

• the result is zero: \( \int_{-a}^{a} f(x) \, dx = 0 \).

This property can greatly simplify our calculations. In our problem, the integrand is an odd function over a symmetric interval from \( k-r \) to \( k+r \).
Therefore, the integral over the entire interval from \( k-r \) to \( k+r \) can be ignored as it evaluates to zero. As a result, we are left focusing mainly on half of the interval for calculations.

Integral Calculus

Integral calculus is a fundamental tool for calculating quantities like fluid force for submerged objects. By integrating the pressure over the surface area, we determine the total fluid force applied to that surface.

• In essence, it involves slicing the surface into infinitely small pieces, calculating the force on each piece, and summing these forces.

For the circular plate example, we calculate the fluid force using the integral:

• \( F = \int_{k-r}^{k+r} w(k-y) 2\pi \sqrt{r^2 - (y-k)^2} \, dy \).

However, by using the properties of odd functions, the integral simplifies significantly.
These simplifications can then lead to evaluating the fluid force by geometric formulas or other techniques.
Integral calculus, therefore, not only helps in setting up the problem but also in finding ways to solve it efficiently.