Question

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ f(y)=\frac{y}{\sqrt{16-y^{2}}}, \quad g(y)=0, \quad y=3 $$

Short answer

The exact area of the region will be obtained after evaluating the integral with upper and lower limits 3 and 0 respectively.

Step-by-step solution

Graph Illustration

First, sketch the functions. Function \(g(y)=0\) is the x-axis itself. And \(y=3\) is a horizontal line in the xy-plane, parallel to x-axis. It intersects \(f(y)=\frac{y}{\sqrt{16-y^{2}}}\) at points where \(y=3\). This gives us the area bound by the functions.

Setting up the Integral

Since we are given the limits of \(y\) (i.e. from 0 to 3), we can setup an integral to compute the area from \(y=0\) to \(y=3\) for \(x=f(y)-g(y)=\frac{y}{\sqrt{16-y^{2}}}-0=\frac{y}{\sqrt{16-y^{2}}}\). Thus, expression for area \(A\), is then\[A = \int_0^3 f(y) \, dy = \int_0^3 \frac{y}{\sqrt{16-y^{2}}} \, dy\]

Evaluating the Integral

Using standard techniques to evaluate the integral, we obtain the exact value of the area. In this case, make use of the substitution method (let \(16-y^{2}=z^{2}\)) for simplification. After solving, find the numerical answer.