Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) $$ \begin{array}{l} f(x, y)=10+2 x-3 y \\ R=\left\\{(x, y): x^{2}+y^{2} \leq 9\right\\} \end{array} $$

Short answer

The area of the surface given by \(z=f(x, y) = 10+2x-3y\) over the region \(R=\left\{(x, y): x^{2}+y^{2} \leq 9\right\}\) is \(9\sqrt{14} \pi \) square units.

Step-by-step solution

Find the partial derivatives

The function \(f(x, y)\) is \(f(x, y)=10+2x-3y\) The partial derivative of this function with respect to x is \(f_{x}=2\). Similarly, the partial derivative of this function with respect to y is \(f_{y}=-3\).

Formulate the surface area formula

The surface area formula requires \(\sqrt{1+(\partial f/\partial x)^{2}+(\partial f/\partial y)^{2}}\). Substituting the values from Step 1, this part of the formula becomes \(\sqrt{1+(2)^{2}+(-3)^{2}}=\sqrt{1+4+9}=\sqrt{14}\).

Transform to polar coordinates

Given that R is described by a disc where \(x^{2}+y^{2} \leq 9\), polar coordinates are appropriate. So, \(x=r \cos(\theta)\) and \(y=r \sin(\theta)\) where \(r \ge 0\) and \(0 \le \theta \le 2 \pi\). \(dx dy\) becomes \(r dr d\theta\). Also in polar coordinates, the area of R ranges from 0 to 3 for the radius (r), and from 0 to \(2 \pi\) for the angle \(\theta\).

Calculate the double integral

Using the surface area formula, \(A=\iint_{R}\sqrt{14} dx dy \), after switching to polar coordinates this integral becomes \(A= \int_{0}^{2 \pi} \int_{0}^{3} \sqrt{14} *r dr d\theta\). So, \(A= \sqrt{14} \int_{0}^{2 \pi} d\theta [\frac{1}{2} r^{2}]_{0}^{3} = \sqrt{14} * 2\pi *[\frac{9}{2} - 0] = 9\sqrt{14} \pi \).

Key concepts

Partial Derivatives

Partial derivatives are essential in the realm of multivariable calculus, as they demonstrate how a function changes as you vary only one of its input variables, while holding the others constant. When we talk about the surface area of a graph defined by a function of two variables, like the function at hand, which is expressed as the equation f(x, y) = 10 + 2x - 3y, we need to understand how the surface bends and twists in 3D space.

To express this numerically, we find the rates of change in both the x and y directions - these are the partial derivatives. For our function, the partial derivatives f_x and f_y are constants, 2 and -3 respectively. These constants tell us that our surface has a uniform linear slant in both the x and y directions. In a more intricate scenario, where the partial derivatives are not constants, the surface might exhibit a more complex behavior, curving and twisting unpredictably.

Applying this concept to the surface area calculation, we use the partial derivatives in the formula to account for the additional area contributed by the incline of the surface. The resulting expression sqrt{1 + (f_x)^2 + (f_y)^2} represents a stretched version of the original flat region 'R', now accounting for the change in height due to the function's slope.

Polar Coordinates

Polar coordinates are incredibly useful, especially when dealing with areas that have circular symmetry, like the disk described in region 'R'. Unlike the rectangular XY plane, polar coordinates use a radius and an angle to specify locations.

To visualize this, picture a point moving away from the origin: the distance from the origin is the radius r, and the angle θ from the positive x-axis gives the direction. This system is perfect for our disk region because it simplifies the area's otherwise complex boundary in Cartesian coordinates into a simple range for r and θ.

In practice, when we transform a function from Cartesian to polar coordinates, we replace x with r*cos(θ) and y with r*sin(θ). We also change the small area element from dx dy to r dr dθ, because as we move outwards from the origin, the slices of area become larger – much like the slices of a cake as you cut further from the center. This is why the factor r is so significant in our double integral; it scales the area to match the geometry of our polar space.

Double Integral

The double integral is a powerful tool for calculating the area under a surface in a two-dimensional region. It's like adding up an infinite number of infinitesimally thin strips of area. But when we talk about surface areas, the double integral does even more - it combines the tiny pieces of surface area over a region R, taking into account the surface's slope.

For our surface area problem, the double integral sums up all the 'stretched' pieces of the region R, which we calculated using the surface area formula including partial derivatives. By substituting the constants obtained from the partial derivatives and transforming the Cartesian region into polar coordinates, we simplify the computation significantly.

The beauty of using polar coordinates here lies in the simplicity it brings to the double integral. Where the bounds are circular, the double integral is often easier to evaluate when expressed in terms of r and θ. This is exactly what we did, calculating the integral over a disk by integrating radially from 0 to 3 (the radius of the disk) and then rotating from 0 to 2π radians. Simplifying the math gives us the final surface area in terms of π, cleanly encapsulating the symmetry and simplicity of the circular shape we started with.