Question

Building Design \(\quad\) The ceiling of a building has a height above the floor given by \(z=20+\frac{1}{4} x,\) and one of the walls follows a path modeled by \(y=x^{3 / 2}\). Find the surface area of the wall if \(0 \leq x \leq 40\). (All measurements are given in feet.)

Short answer

The solution involves calculating the length of the two curves and deducting one from the other. Performing the proper calculations using the calculus tools will yield the solution.

Step-by-step solution

Define the curves

The curved wall follows the path \(y=x^{3 / 2}\), and the ceiling is defined by the curve \(z=20+\frac{1}{4} x\). These two curves intersect at \(x = 0\) and \(x = 40\).

Calculate the lengths of the curves

To achieve this, one needs to calculate the derivative of the two curves. For the wall, \( y' = \frac{3}{2}*x^{1/2}\) and for the ceiling, \( z' = 1/4 \). Afterward, use these derivatives in the length formula \(\int_0^{40} \sqrt{1 + (y')^2}dx\) and \(\int_0^{40} \sqrt{1 + (z')^2}dx\), respectively.

Subtract to get the length of the wall

To get the length of the wall, subtract these two curves from each other, \(\int_0^{40} \sqrt{1 + [\frac{3}{2}*x^{1/2}]^2}dx - \int_0^{40} \sqrt{1 + (1 / 4)^2}dx\). This can be performed using standard calculation routines or with the help of software tools if needed or applicable.