Problem 29 - College Algebra and Calculus: An Applied Approach

Question

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(y)=\sqrt{y}, y=9, x=0 $$

Step-by-Step Solution

Answer

The area of the bounded region is 18 \(units^2\).

Step 1: Understand and plot the functions

The functions are \(f(y)=\sqrt{y}\), \(y=9\), and \(x=0\). Here, \(f(y)=\sqrt{y}\) is a half-parabola opening rightward. \(y=9\) is a horizontal line, and \(x=0\) is the y-axis. Plot them on a graph and identify the bounded region.

Step 2: Set up the integral

Since the boundaries are described in terms of \(y\), we'll solve this in terms of \(y\). The bounded region is from \(y=0\) to \(y=9\). Hence, the integral setup to find the area will be:\[ \int_0^{9} (x_{right} - x_{left}) \, dy\]Here, \(x_{right}= f(y) = \sqrt{y}\) and \(x_{left} = 0\]Simplifying, the integral becomes:\[ \int_0^{9} (\sqrt{y} - 0) \, dy\]

Step 3: Calculate the area

The area \(A\) of the region can be calculated by performing the integration:\[A = \int_0^{9} \sqrt{y} \, dy = [\frac{2}{3} y^{\frac{3}{2}}]_0^{9} = \frac{2}{3}(9)^{\frac{3}{2}} - \frac{2}{3}(0)^{\frac{3}{2}} = 18 \, units^2\]