Problem 12 - Thomas Calculus

Question

a. Polar coordinates \(\quad\) Show, by changing to polar coordinates, that $$\int_{0}^{a \sin \beta} \int_{y \cot \beta}^{\sqrt{a^{2}-y^{2}}} \ln \left(x^{2}+y^{2}\right) d x d y=a^{2} \beta\left(\ln a-\frac{1}{2}\right)$$ where \(a>0\) and \(0<\beta<\pi / 2\) b. Rewrite the Cartesian integral with the order of integration reversed.

Step-by-Step Solution

Answer

The given integral equals \( a^2 \beta (\ln a - \frac{1}{2}) \) in polar coordinates; reversing integration gives \( \int_{0}^{a} \int_{x \cot \beta}^{\sqrt{a^2 - x^2}} \ln(x^2 + y^2) \, dy \, dx \).

Step 1: Express Integral in Polar Coordinates

To express the integral in polar coordinates, replace Cartesian coordinates \((x, y)\) with polar coordinates \((r, \theta)\) using the transformations \( x = r \cos\theta \) and \( y = r \sin\theta \). In polar coordinates, the Jacobian for the transformation is \( r \), so the differential area is \( dx\,dy = r\,dr\,d\theta \).

Step 2: Identify Limits of Integration

The original integral limits \( 0 \leq y \leq a \sin \beta \) and \( y \cot \beta \leq x \leq \sqrt{a^2 - y^2} \) in Cartesian coordinates need to be identified in polar coordinates. The curve \( y = a \sin \beta \) becomes \( r \sin \theta = a \sin \beta \) or \( r = \frac{a \sin \beta}{\sin \theta} \), which is limited by \( \theta = 0 \) to \( \theta = \beta \). Thus, the region is bounded by \( 0 \leq r \leq a \), and \( 0 \leq \theta \leq \beta \).

Step 3: Transform the Integral

Transform the given integral into polar coordinates:\[\int_{0}^{\beta} \int_{0}^{a} \ln(r^2) \cdot r \, dr \, d\theta.\]The \( \ln(x^2 + y^2) \) term converts into \( \ln(r^2) \). The integral becomes separable into \( \int_{0}^{a} \ln(r^2) r \, dr \) and \( \int_{0}^{\beta} \, d\theta \).

Step 4: Solve the Integral

First, solve the \( r \) integral:\[\int_{0}^{a} r \ln(r^2) \, dr = \int_{0}^{a} 2r \ln(r) \, dr.\]Using integration by parts, let \( u = \ln(r) \) and \( dv = 2r \, dr \), so \( du = \frac{1}{r} \, dr \) and \( v = r^2 \):\[\begin{aligned}\int 2r \ln(r) \, dr & = r^2 \ln(r) - \int r^2 \, dr \& = [r^2 \ln(r) - \frac{r^3}{3}]_{0}^{a} = [a^2 \ln(a) - \frac{a^3}{3}] - [0] \& = a^2 \ln(a) - \frac{a^3}{3}.\end{aligned}\]Then multiply by \( \int_{0}^{\beta} d\theta = \beta \):\[\beta \left(a^2 \ln(a) - \frac{a^3}{3}\right).\]

Step 5: Calculate the Final Result

Combine the terms:\[a^2 \beta (\ln(a) - \frac{1}{2})\]This confirms the given solution of the integral in polar coordinates, as required.

Step 6: Reverse the Order of Integration

For reversing the order of integration, first analyze the region of integration in the \( xy \)-plane:- The original limits were \( 0 \leq y \leq a \sin \beta \) and \( y \cot \beta \leq x \leq \sqrt{a^2 - y^2} \).- The interval for \( x \) from \( 0 \) to \( a \) is divided depending on \( y \)'s limit.Now, the reversed order of the region involves fixing \( x \) first:- \( 0 \leq x \leq a \).- For each \( x \), \( x \cot \beta \leq y \leq \sqrt{a^2 - x^2} \).Thus, the integral becomes:\[\int_{0}^{a} \int_{x \cot \beta}^{\sqrt{a^2 - x^2}} \ln(x^2 + y^2) \, dy \, dx.\]

Key Concepts

Integration by Parts

Integration by Parts is a powerful technique from calculus that helps in solving integrals where the integrand is a product of two functions. The idea is based on the product rule of differentiation and is especially useful when one part of the function becomes simpler upon differentiation, and the other part upon integration. The formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

To apply this technique, we select which part of the integrand to set as \( u \) and which part as \( dv \). The differentiation of \( u \) yields \( du \), and integration of \( dv \) provides \( v \). In the exercise, to simplify the integral \( \int 2r \ln(r) \, dr \), we choose \( u = \ln(r) \) and \( dv = 2r \, dr \). This choice helps in reducing the function’s complexity through differentiation, as \( \ln(r) \) becomes \( \frac{1}{r} \). By substituting back into the integration by parts formula, it allows the integral to be more manageable. Concluding the integration step with the initial conditions aids in obtaining the final expression.

Coordinate Transformation

When dealing with integrals over non-rectangular domains, transforming from Cartesian coordinates (\( x, y \)) to Polar coordinates (\( r, \theta \)) can greatly simplify the evaluation process. The relationship between these two coordinate systems is given by:

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

The change also requires account for the differential area transformation, where \( dx \ dy = r \, dr \, d\theta \). This factor, \( r \), acts as the Jacobian determinant of the transformation from Cartesian to Polar coordinates. Such transformations enable more straightforward integration, particularly for circular or radial symmetrical regions. In this exercise, transforming the integrand \( \ln(x^2 + y^2) \) converts into \( \ln(r^2) \), simplifying due to radial symmetry. Additionally, defining radius \( r \) and angle \( \theta \) bound the domain more intuitively by using the expressions derived from the limits in Cartesian coordinates, allowing easier computation of the integral over this transformed domain.

Reversing Integration Order

Reversing the order of integration can be particularly useful when the original region of integration is complex or when the inner integral is hard to tackle within its current limits. The approach involves reevaluating the limits of integration for the variables involved.First, visualize the region defined by the integration limits. In this case, the region in the xy-plane is initially bound by:

• \( 0 \leq y \leq a \sin \beta \)
• \( y \cot \beta \leq x \leq \sqrt{a^2 - y^2} \)

By observing this region, we can switch the order of integration to simplify computation. Here, fixing \( x \) first by setting limits for \( y \) over a constant \( x \) range transforms the bounds into:

• \( 0 \leq x \leq a \)
• \( x \cot \beta \leq y \leq \sqrt{a^2 - x^2} \)

Such rearrangement can often lead to integrals that are easier to evaluate, due to simpler boundaries or easier integrands in the reversed order. It provides flexibility in the integration process, allowing us to handle more complicated regions by selecting the most efficient computation path.