Problem 14 - APEX Calculus

Question

In Exercises \(11-14,\) an iterated integral in rectangular coordinates is given. Rewrite the integral using polar coordinates and evaluate the new double integral. $$ \begin{array}{l} \int_{-2}^{-1} \int_{0}^{\sqrt{4-x^{2}}}(x+5) d y d x+\int_{-1}^{1} \int_{\sqrt{1-x^{2}}}^{\sqrt{4-x^{2}}}(x+5) d y d x+ \\ \int_{1}^{2} \int_{0}^{\sqrt{4-x^{2}}}(x+5) d y d x \end{array} $$

Step-by-Step Solution

Answer

The integral in polar coordinates evaluates to \(\frac{15\pi}{2}\).

Step 1: Understand the Region of Integration

The given iterated integrals represent the area on the coordinate plane we need to integrate over. It involves a semicircle with radii of 2 (left and right limits) and a semicircle of radii 1 (top and bottom limits) centered at the origin. Our integration goes across this region in two parts: \(-2 \leq x \leq -1\), \(-1 \leq x \leq 1\), and \(1 \leq x \leq 2\). The function is complicated between the smaller and larger semicircles.

Step 2: Convert Bounds to Polar Coordinates

In polar coordinates, a circle with radius \(r\) is defined as \(r^2 = x^2 + y^2\). The angle \(\theta\) varies from 0 to \(\pi\) (for the upper semicircle). We find that for \(r\) goes from 1 to 2.

Step 3: Express Function in Polar Coordinates

Convert the function \(x + 5\) into polar coordinates. Since \(x = r\cos(\theta)\), the function can be written as \(r\cos(\theta) + 5\).

Step 4: Rewrite the Integral

The given iterated integral in rectangular coordinates transforms into a polar integral: \[\int_0^{\pi} \int_1^2 (r \cos(\theta) + 5) \cdot r \, dr \, d\theta.\] The additional \(r\) comes from the Jacobian determinant when converting areas from rectangular to polar.

Step 5: Evaluate the Integral with Respect to \(r\)

Integrate with respect to \(r\): \[\int_0^{\pi} \left[\frac{r^3}{3}\cos(\theta) + 5\frac{r^2}{2}\right]_1^2 \, d\theta.\] Substituting the bounds 1 and 2 for \(r\) gives: \[\int_0^{\pi} \Bigg[\frac{8}{3}\cos(\theta) + 10 - \left(\frac{1}{3}\cos(\theta) + \frac{5}{2}\right)\Bigg] \, d\theta.\] This simplifies to \[\int_0^{\pi} \left(\frac{7}{3}\cos(\theta) + \frac{15}{2}\right) \, d\theta.\]

Step 6: Evaluate the Integral with Respect to \(\theta\)

Integrate with respect to \(\theta\): \[\left[\frac{7}{3}\sin(\theta) + \frac{15}{2}\theta\right]_0^{\pi}.\] Since \(\sin(0) = \sin(\pi) = 0\), substitute to find: \[\frac{15}{2}(\pi - 0) = \frac{15\pi}{2}.\]

Step 7: Combine Results and Finalize Answer

Conclude that the evaluation of the given integral, after conversion to polar coordinates, yields \(\frac{15\pi}{2}\).

Key Concepts

Iterated Integral

An iterated integral involves integrating a function over a certain region in the coordinate plane. Typically, this means performing one integration after another, such as integrating first with respect to one variable and then with respect to another. For example, if you have a function \( f(x,y) \), an iterated integral might look like \( \int_a^b \int_c^d f(x,y) \, dy \, dx \), where \( dy \) is integrated first, followed by \( dx \). The order of integration can sometimes be changed, but care must be taken with the limits of integration. Iterating in a different order can sometimes make an integral easier to solve. In practice, iterated integrals are essential for computing areas and volumes in different coordinate systems.

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, represent any point in a plane using an \( (x, y) \) pairing. Here, \( x \) signifies the horizontal distance from the origin, and \( y \) represents the vertical distance. This coordinate system is intuitive and linear, which makes it very useful for many mathematical operations involving straight lines or rectangular regions. In the context of iterated integrals, these coordinates often serve as a starting point for setting up the integrals over specific regions. However, regions bounded by curves such as circles can be more efficiently handled by converting to other coordinate systems like polar coordinates.

Double Integral

A double integral is an extension of the concept of a single integral to functions of two variables. It helps compute the volume under the surface defined by \( z = f(x, y) \) over a region on the \( xy \)-plane. In a double integral, such as \( \int \int f(x, y) \, dx \, dy \), you're performing the integral operation twice — once for each of the two variables. This allows us to find a cumulative sum over a two-dimensional area. Double integrals can be simplified or evaluated more conveniently by switching from rectangular to polar coordinates, especially when dealing with circular symmetry. This switch is facilitated through an additional factor known as the Jacobian determinant, which accounts for the change of area when coordinates are transformed.

Jacobian Determinant

When converting between coordinate systems, such as from rectangular to polar coordinates, the Jacobian determinant is a crucial factor. It represents how volumes or areas change during the transformation. For the conversion to polar coordinates, the Jacobian is \( r \). This factor scales the differential area \( dA = dx \, dy \) to \( dA = r \, dr \, d\theta \) in polar coordinates, ensuring the integral remains accurate. Without incorporating the Jacobian determinant, the calculated areas or volumes could be incorrect. The expression \( r \, dr \, d\theta \) adjusts for the how space is "stretched" or "squeezed" by the transformation. This step is vital when evaluating integrals involving radial symmetry or circular regions, making the calculations significantly easier and more intuitive.