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(k>r)feet below the surface of the fluid. Show that the fluid force on the surface of the plate is F=wk(πr^2). How to prove this given formula by Evaluating one integral by a geometric formula and the other by observing that the integral is an odd function?

To prove the formula for the fluid force on the surface of the plate, we can use the concept of pressure and integrate it over the surface of the plate.

1. First, let's consider a small area element on the surface of the circular plate. The force on this small element is determined by the pressure exerted by the fluid. The pressure at a depth h feet below the surface of the fluid is given by:

P = wh,

where w is the weight of the fluid per cubic foot.

2. The force on the small area element can be calculated by multiplying the pressure by the area of the element. Since the small area element is nearly perpendicular to the fluid surface, we can consider the force acting normal to the surface.

dF = P * dA,

where dA is the differential area element on the surface of the plate.

3. To express dA in terms of the variable of integration, we can consider the area of a differential strip on the surface of the plate. Let's consider a strip at a distance y from the center, with a width dy. The length of this strip is given by 2√(r^2 - y^2), where r is the radius of the circular plate.

dA = 2√(r^2 - y^2) * dy.

4. Now, let's integrate the force over the entire surface of the plate by integrating dF over the appropriate limits. Since the strip starts at y = k - r and ends at y = k, the integration limits are (k - r) to k.

F = ∫[(k - r) to k] P * dA.

5. Substituting the expressions for P and dA, we have:

F = ∫[(k - r) to k] wh * (2√(r^2 - y^2) * dy).

6. Simplifying the equation, we have:

F = 2w * ∫[(k - r) to k] √(r^2 - y^2) * dy.

7. Since the integral involves the radical function, we can use a geometric formula to evaluate it. The integral:

∫ √(r^2 - y^2) * dy

represents the area of a semicircle with radius r. So, the integral becomes:

∫ √(r^2 - y^2) * dy = (π/2)*r^2.

8. By substituting this result back into the earlier equation, we get:

F = 2w * (π/2) * r^2 = wπr^2.

9. Finally, multiplying the force by w to incorporate the weight of the fluid, we have:

F = w * wπr^2 = wkπr^2.

Therefore, we have proved that the formula for the fluid force on the surface of the circular plate is F = wkπr^2.

In this proof, we evaluated one integral using a geometric formula (the area of a semicircle) and observed that the other integral is an odd function, meaning that it integrates to zero over the symmetric interval (k - r) to k.