Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) $$ \begin{array}{l} f(x, y)=\sqrt{a^{2}-x^{2}-y^{2}} \\ R=\left\\{(x, y): x^{2}+y^{2} \leq a^{2}\right\\} \end{array} $$

Short answer

The area of the surface given by \(z=f(x, y)=\sqrt{a^{2}-x^{2}-y^{2}}\) over the region \(R = \left\{ (x, y) : x^{2}+y^{2}\leq a^{2} \right\}\) is \(\pi a^{2}\), which corresponds to the surface area of a hemisphere of radius \(a\).

Step-by-step solution

Understand the Geometry of the Problem

Recognize that the region \(R\) is a disk of radius \(a\) in the \(xy\)-plane. Also, the surface is a hemisphere (from the equation of \(f(x, y)\)) of radius \(a\) above the \(xy\)-plane.

Compute the Integrals

The formula for the surface area of a function \(z=f(x, y)\) over a region \(R\) in the \(xy\)-plane is given by \(A=\int \int_{R} \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}} d x d y\). Here, both \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) equal \(-\frac{x}{\sqrt{a^{2}-x^{2}-y^{2}}}\) and \(-\frac{y}{\sqrt{a^{2}-x^{2}-y^{2}}}\) respectively. Plugging these into the formula simplifies it to \(A=\int \int_{R} d x d y\), because \(\sqrt{1+\left(\frac{x}{\sqrt{a^{2}-x^{2}-y^{2}}}\right)^{2}+\left(\frac{y}{\sqrt{a^{2}-x^{2}-y^{2}}}\right)^{2}}=\sqrt{1+\frac{x^{2}+y^{2}}{a^{2}-x^{2}-y^{2}}\}=1\).

Calculate Area

To calculate the area \(A=\int \int_{R} d x d y\) over the region \(R=\left\{(x, y): x^{2}+y^{2} \leq a^{2}\right\}\), switch to polar coordinates. This is encouraged because the region \(R\) is a disk centered at the origin, and the area element in polar coordinates is \(dx dy=r dr d\theta\). Make the switch, and compute the double integral \(A=\int_{0}^{2 \pi} \int_{0}^{a} r d r d \theta\) to determine the surface area.