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Chapter 11: Problem 29

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

Short Answer

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

Step by step solution

01

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.
02

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:\[ \int_0^{9} (\sqrt{y} - 0) \, dy\]
03

Calculate the area

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

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