Problem 14 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The value of the polar integral is \(\frac{1}{15}\).

Step 1: Understand the Region of Integration

The given integral is in Cartesian coordinates and ranges over a specific region in the xy-plane. It integrates from \(x = -\sqrt{1-(y-1)^{2}}\) to \(x=0\) and from \(y=0\) to \(y=2\). This region describes a semi-circle with center at \((0, 1)\) and radius 1, specifically the left half, and the line dividing the region at the y-axis (\(x=0\)).

Step 2: Convert Cartesian Region to Polar Coordinates

In polar coordinates, the transformations \(x = r \cos(\theta)\), \(y = r \sin(\theta)\) are used, and the Jacobian \(r\) is multiplied in the integrand. The semi-circle is expressed as \(0 \leq r \leq 1\), and the limits for \(\theta\) are \(\pi/2 \leq \theta \leq \pi\) to account for the left half of the circle.

Step 3: Convert the Integrand from Cartesian to Polar

The integrand \(xy^2\) translates to \((r \cos(\theta))(r^2 \sin^2(\theta)) = r^3 \cos(\theta) \sin^2(\theta)\). Multiplying by the Jacobian \(r\), the integrand becomes \(r^4 \cos(\theta) \sin^2(\theta)\).

Step 4: Set Up the Polar Integral

The Cartesian integral \(\int_{0}^{2} \int_{-\sqrt{1-(y-1)^{2}}}^{0} xy^2 \, dx \, dy\) is converted to the polar integral \(\int_{\pi/2}^{\pi} \int_{0}^{1} r^4 \cos(\theta) \sin^2(\theta) \, dr \, d\theta\).

Step 5: Evaluate the Inner Integral

Compute the inner integral with respect to \(r\): \[\int_{0}^{1} r^4 \, dr = \left[\frac{r^5}{5}\right]_0^1 = \frac{1}{5}.\]

Step 6: Evaluate the Outer Integral

Substitute the result of the inner integral into the outer integral and compute it: \[\int_{\pi/2}^{\pi} \cos(\theta) \sin^2(\theta) \, d\theta = \frac{1}{5} \int_{\pi/2}^{\pi} \cos(\theta) \sin^2(\theta) \, d\theta.\] Using the substitution \( u = \sin(\theta)\), \(du = \cos(\theta) \, d\theta\), transform limits from \(\pi/2\) to 0 and \(\pi\) to 1:\[\int_{0}^{1} u^2 \, du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3}.\]Thus, the outer integral evaluates to \(\frac{1}{5} \cdot \frac{1}{3} = \frac{1}{15}.\)

Step 7: Conclusion of Evaluation

The final evaluation of the polar integral gives the value \(\frac{1}{15}\).

Key Concepts

Cartesian Integral

To begin understanding this problem, it's crucial to grasp the concept of a Cartesian integral. In Cartesian coordinates, functions are integrated over a specified region in the xy-plane. For example, the integral \( \int_{0}^{2} \int_{-\sqrt{1-(y-1)^{2}}}^{0} x y^{2} \, dx \, dy \) is evaluated over a region defined by the integration limits for \(x\) and \(y\). In this instance, we are integrating over a semi-circle centered at \((0, 1)\) in the left half of the plane and it divides the region at the y-axis \((x=0)\). This forms the basis for translating the problem into different coordinate systems, such as polar coordinates, to potentially simplify computations. To convert this Cartesian integral into a more manageable form, we move towards understanding polar integrals.

Polar Integral

Converting a Cartesian integral into a polar integral involves transforming both the region of integration and the integrand. In polar coordinates, every point in the plane is described by a distance \(r\) from the origin and an angle \( \theta \) from the positive x-axis. For this conversion, use the equations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Additionally, the Jacobian determinant \( r \) is applied to adjust the integrand for area element size change.
For the integral \( \int_{0}^{2} \int_{-\sqrt{1-(y-1)^{2}}}^{0} x y^{2} \, dx \, dy \), we translate it into polar coordinates as \( \int_{\pi/2}^{\pi} \int_{0}^{1} r^{4} \cos(\theta) \sin^{2}(\theta) \, dr \, d\theta \). This reflects a semi-circle with radius 1, centered at \((1, 0)\), for angles from \( \pi/2 \) to \( \pi \). Now, solving the problem involves evaluating this polar-formulated integral.

Coordinate Transformation

Coordinate transformation is a pivotal step in changing the system of integration from Cartesian to polar. The transformation helps in redefining the integration boundaries and the integrand function for ease of integration. In the provided example, the Cartesian limits are initially rectangular but instead of directly handling complex limits, transforming to polar coordinates outlines a simpler region, specifically a half-circle.
Transform points by the equations:

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

Thus the complex Cartesian boundaries involving square roots become simple radial sweeps in polar space. This transformation simplifies evaluating integrals significantly, especially when dealing with circular or radial symmetry.

Jacobian in Integration

The Jacobian is a critical component in multi-variable calculus that scales the area elements appropriately during a transformation of coordinates. When converting integrals from Cartesian to polar coordinates, the Jacobian \( r \) arises from the determinant of the transformation matrix involving partial derivatives of \(x\) and \(y\) with respect to \(r\) and \(\theta\).
Using the transformation equations:

• \( dx \ dy = r \ dr \ d\theta \)

This addition of \( r \) is crucial because it accounts for the distortion of space when moving from a grid in Cartesian space to a circular grid in polar space. The Jacobian ensures that every infinitesimal area in the new coordinate system accurately represents the same physical space as the original. Thus, it maintains the integral's value while simplifying the calculation process across more easily managed regions.