Question

$$ f(x, y)=\arctan \frac{y}{x}, R: x^{2}+y^{2} \geq 1, x^{2}+y^{2} \leq 4,0 \leq y \leq x $$$$ f(x, y)=9-x^{2}-y^{2}, R: x^{2}+y^{2} \leq 9, x \geq 0, y \geq 0 $$

Short answer

These problems involve visualizing the given functions in their respective regions. The first function, \(f(x, y)=\arctan \frac{y}{x}\), is represented in a region between two concentric circles where y is greater than or equal to 0 and less than or equal to x. The function would generally depict smooth changes across the region, symmetric to the line y=x. The second function, \(f(x, y)=9-x^{2}-y^{2}\) is visualized within a quarter circle in the first quadrant of the xy-plane. It is a downward opening paraboloid symmetric across the line y=x, and decreases as you move away from the origin.

Step-by-step solution

Visualize first function within its region

For the first function \(f(x, y)=\arctan \frac{y}{x}\) in the region \(R: x^{2}+y^{2} \geq 1, x^{2}+y^{2} \leq 4,0 \leq y \leq x\), one needs to sketch the region R in the xy-plane. Plot two concentric circles, one with radius 1 and one with radius 2. Only consider points within these two circles where y is greater than or equal to 0 and less than or equal to x. Now visualize the function \(f(x,y)\) within this region. As it's a trigonometric function involving inverse tangent, we can expect that it will change smoothly as x and y change within the given region and that it will be symmetric about the line y = x.

Visualize second function within its region

The second function is \(f(x, y)=9-x^{2}-y^{2}\) and the region \(R: x^{2}+y^{2} \leq 9, x \geq 0, y \geq 0\). Once again, sketch the region R, which is now a quarter of a circle with radius 3 in the first quadrant of xy-plane. This function is a downward-opening paraboloid and you can see that it will be symmetric across the line y = x, and decreasing as you move away from the origin.

Key concepts

Double Integration

Double integration is a powerful tool for calculating the volume under a surface in a specified region. It involves integrating a function of two variables, say, \( f(x, y) \) over a region \( R \) in the xy-plane. The process is akin to summing up infinitesimally small rectangles with heights given by the function value and area \( dx \cdot dy \) over the entire region.

In the given exercise, double integration can be applied to find the volume under the surfaces defined by \( f(x, y) = \arctan \frac{y}{x} \) and \( f(x, y) = 9 - x^2 - y^2 \) within their respective regions. The integration bounds are derived from the inequalities describing the region \( R \). For instance, the region for the first function is a set of points between two circles, accounting for the limits on \( x \) and \( y \) as given. To compute the volume, you would set up the double integral with the appropriate limits and integrate the function \( f \) with respect to \( y \) and then \( x \) (or vice versa, depending on the specifics of the region and integration order).

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction. The reference point is known as the pole, and the reference direction typically points to the right from the pole, following the positive x-axis. A point in the polar coordinate system is represented by \( (r, \theta) \), where \( r \) is the radial distance from the pole, and \( \theta \) is the polar angle from the reference direction.

Converting to polar coordinates is particularly helpful when dealing with regions bounded by circles or when the integrand and limits have a clear radial symmetry. For the given problem's regions, identified by circles, polar coordinates simplify the double integration process. To convert, use \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \)). This transformation also requires updating the differential area element from \( dx dy \) to \( r dr d\theta \) due to the Jacobian determinant when changing variables in integration.

Volume of Solids

The volume of a solid bounded by a surface in three dimensions can be determined using double integrals. By integrating a surface function \( f(x, y) \) over its projected area on the xy-plane, we obtain the volume of the solid that lies underneath the surface and above the plane within a given boundary.

For the exercise problems at hand, the volumes of solids defined above the xy-plane and beneath the graphs of \( f(x, y) = \arctan \frac{y}{x} \) and \( f(x, y) = 9 - x^2 - y^2 \) in their respective regions can be calculated using double integration. When setting up the double integral for volume, one must be aware of the region over which the integration is to be performed and the orientation of the solid. The selection of vertical or horizontal slices can impact the setup of your integration limits and, at times, the choice of integrating in Cartesian versus polar coordinates.

Trigonometric Functions

Trigonometric functions like \( \sin \), \( \cos \), and \( \tan \) describe relationships between the angles and sides of a right-angled triangle. Their inverses, such as \( \arcsin \), \( \arccos \), and \( \arctan \), are used to find the angles given the sides' ratios. These functions are periodic and exhibit symmetry, which are essential traits that can immensely simplify integration, especially in polar coordinates.

In our first example function, \( f(x, y) = \arctan \frac{y}{x} \) is an inverse trigonometric function yielding angles and can affect the strategy for integration. Depending on the region of integration, recognizing symmetries due to trigonometric functions can allow for the simplification of the integral, such as integrating over just a portion of the region and then multiplying by a symmetry factor, although care must be taken to apply this method correctly depending upon the function's behavior over the region.