Problem 21 - University Calculus: Early Transcendentals

Question

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$\int_{0}^{1} \int_{x}^{\sqrt{2-x^{2}}}(x+2 y) d y d x$$

Step-by-Step Solution

Answer

The equivalent polar integral evaluates to \(\frac{8\sqrt{2} - 2}{3}.\)

Step 1: Understand the region of integration

The Cartesian integral is given by \( \int_{0}^{1} \int_{x}^{\sqrt{2-x^{2}}}(x+2y) \ dy \ dx \). The limits for \(y\) range from \(y = x\) to \(y = \sqrt{2-x^2}\). The limits for \(x\) range from \(x = 0\) to \(x = 1\). We need to identify the region in the Cartesian plane that these bounds define.

Step 2: Sketch the region and convert boundaries to polar coordinates

In the Cartesian plane, the boundary \(y = x\) is a line, and \(y = \sqrt{2 - x^2}\) represents the upper half of a circle with radius \(\sqrt{2}\). Therefore, the region is between this half-circle and the line from \((0,0)\) to \((1,1)\). In polar coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\). The circle has the equation \(r^2 = 2\), and \(\theta\) varies from \(\pi/4\) to \(\pi/2\).

Step 3: Convert the integral to polar coordinates

Using the relations \(x = r\cos\theta\), \(y = r\sin\theta\), and the Jacobian determinant \(r\), the integral becomes \(\int_{\pi/4}^{\pi/2} \int_{0}^{\sqrt{2}} (r \cos\theta + 2r^2 \sin\theta) r dr d\theta\). Simplify to get: \(\int_{\pi/4}^{\pi/2} \int_{0}^{\sqrt{2}} (r^2 \cos\theta + 2r^3 \sin\theta) dr d\theta\).

Step 4: Integrate with respect to \(r\)

Evaluate the integral \(\int_{0}^{\sqrt{2}} (r^2 \cos\theta + 2r^3 \sin\theta) dr\). This gives: \[ \left[ \frac{r^3}{3} \cos\theta + \frac{r^4}{2} \sin\theta \right]_{0}^{\sqrt{2}} = \left( \frac{(\sqrt{2})^3}{3} \cos\theta + \frac{(\sqrt{2})^4}{2} \sin\theta \right) - 0 \].

Step 5: Simplify the result of \(r\)-integration

Substitute \(r = \sqrt{2}\): \(\frac{2\sqrt{2}}{3} \cos \theta + 4 \sin \theta\). This is the result of the \(r\)-integration.

Step 6: Integrate with respect to \(\theta\)

Evaluate \(\int_{\pi/4}^{\pi/2} \left( \frac{2\sqrt{2}}{3} \cos \theta + 4 \sin \theta \right) d\theta\). The result is: \[ \left[ \frac{2\sqrt{2}}{3} \sin \theta - 4 \cos \theta \right]_{\pi/4}^{\pi/2} = \left( \frac{2\sqrt{2}}{3}(1) - 4(0) \right) - \left( \frac{2\sqrt{2}}{3} \left(\frac{\sqrt{2}}{2}\right) - 4 \left(\frac{\sqrt{2}}{2}\right) \right) \].

Step 7: Simplify and calculate the final result

Simplify the expression to get: \[ \frac{2\sqrt{2}}{3} - \left( \frac{2}{3} - 2\sqrt{2} \right) = \frac{2\sqrt{2}}{3} - \frac{2}{3} + 2\sqrt{2} = \frac{8\sqrt{2} - 2}{3} = \frac{8\sqrt{2} - 2}{3} \].

Key Concepts

Cartesian to Polar Conversion

When dealing with double integrals in polar coordinates, one often begins by converting a given Cartesian integral to polar form. To understand this conversion, it's essential to recognize the relationship between the two coordinate systems.

The Cartesian coordinates \(x, y\) relate to the polar coordinates \(r, \theta\) through the formulas:

•  \(x = r \cos \theta\)
•  \(y = r \sin \theta\)

In this exercise, the region of integration described by the boundaries \(y = x\) and \(y = \sqrt{2 - x^2}\) maps onto an angular range in polar coordinates. Specifically, for the circle defined by \(r^2 = 2\), the angle \(\theta\) varies between \(\pi/4\) and \(\pi/2\).

This process involves expressing the function in the integral in terms of \(r\) and \(\theta\), and modifying the limits of integration accordingly. The Jacobian, \(r\), emerges in the integration due to the transformation from Cartesian to polar coordinates. After conversion, the integral represents the same area or volume, just expressed in a different coordinate system.

Double Integrals

Double integrals allow you to compute the volume under a surface over a specific region on a plane. To comprehend this, visualize the surface as an umbrella over a patch on the ground, where the integral sums up tiny vertical "slices" over that patch.

In Cartesian coordinates, these integrals are expressed typically as \(\int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx\). This formulation translates into an iterative process:

•  First, integrate with respect to the inner variable.
•  Later, compute the outer integral to accumulate the final result.

Switching to polar coordinates can simplify these integrals when dealing with circular regions.

In the polar form \( \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r, \theta) \, r \, dr \, d\theta\), the extra \(r\) in the integrand (from the Jacobian) adjusts for the radial 'stretching' of area elements. This conversion inherently aligns the method of integration with the symmetry often present in circular regions.

Integration Techniques

Integration is a process of finding the integral of a function, a fundamental concept in calculus. Different techniques are utilized to evaluate integrals effectively, especially in double integrals. For polar coordinates, the strategies align with the symmetry and geometric form of the region of integration.

Start by transforming the Cartesian function into a polar one, as discussed earlier — this often simplifies computations if the region is circular. Then the choice of \(r\) and \(\theta\) limits is adjusted to reflect the problem's symmetry.

In this exercise's solution, the sequence involved:

•  Clearly defining the region to integrate over using appropriate limits.
•  Strategically calculating the inner integral with respect to \(r\), while the outer integral operates over \(\theta\).
•  Finally, refining the result for the \(r\)-integration before evaluating the outer integral.

This systematic and structured approach ensures efficient handling of complex areas, providing insights into effectively calculating integrals over non-rectangular domains.