Problem 11 - University Calculus: Early Transcendentals

Question

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$\int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}}\left(x^{2}+y^{2}\right) d x d y$$

Step-by-Step Solution

Answer

The value of the integral is \(2\pi\).

Step 1: Identify Region of Integration

The given integral is \(\int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} (x^2 + y^2) \, dx \, dy\). The inner integral from \(x = 0\) to \(x = \sqrt{4 - y^2}\) describes a semicircle centered at the origin with radius 2 in the upper half-plane, since \(x = \sqrt{4 - y^2}\) is equivalent to \(x^2 + y^2 = 4; \ x \geq 0\).

Step 2: Convert Cartesian to Polar Coordinates

To convert to polar coordinates, use the transformations \(x = r \cos \theta\) and \(y = r \sin \theta\), and note that \(x^2 + y^2 = r^2\). The Jacobian of the transformation is \(r\). In polar coordinates, the region is described by \(0 \leq r \leq 2\) and \(0 \leq \theta \leq \frac{\pi}{2}\).

Step 3: Set up the Polar Integral

Substitute the polar transformations into the integral: \[\int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^2 \cdot r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \, dr \, d\theta.\]

Step 4: Evaluate the Inner Integral

Evaluate \(\int_{0}^{2} r^3 \, dr\). The antiderivative of \(r^3\) is \(\frac{r^4}{4}\). Thus, \[\int_{0}^{2} r^3 \, dr = \left. \frac{r^4}{4} \right|_{0}^{2} = \frac{16}{4} - 0 = 4.\]

Step 5: Evaluate the Outer Integral

Substitute the result of the inner integral into the outer integral: \[\int_{0}^{\frac{\pi}{2}} 4 \, d\theta.\] The antiderivative of \(4\) with respect to \(\theta\) is \(4\theta\). Thus, \[\left. 4\theta \right|_{0}^{\frac{\pi}{2}} = 4\left( \frac{\pi}{2} \right) = 2\pi.\]

Step 6: Final Result

The value of the original integral, by converting to polar coordinates and evaluating, is \(2\pi\).

Key Concepts

Cartesian to Polar Coordinates

When working with integrals, sometimes it's beneficial to change the coordinate system. One of the most common transformations is from Cartesian to polar coordinates. The Cartesian coordinate system uses the x and y axes to specify any point in the plane. Points are defined in terms of their horizontal and vertical distances away from the origin. In contrast, the polar coordinate system uses a distance from the origin, denoted by \(r\), and an angle \(\theta\) measured from the positive x-axis. The conversion of Cartesian to polar coordinates can be defined as:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(x^2 + y^2 = r^2\)

Switching from Cartesian to polar coordinates is often useful when the region of integration is circular, as was the case in our problem. This method simplifies the calculation by taking advantage of the symmetry of the circle.

Jacobian of Transformation

Whenever you change coordinate systems in an integral, you must account for how area elements change size. This is done using something called the Jacobian of Transformation. This Jacobian is a determinant that accounts for the scaling of area when converting between coordinate systems. In polar coordinates, the differential area element \(dx \, dy\) translates to \(r \, dr \ d\theta\). The term \(r\) is the Jacobian for the polar coordinate transformation.The integration in polar coordinates includes this Jacobian, as it adjusts for the "stretching" that occurs when changing from rectangular to polar coordinates.

• In our example, this means turning the integrand \( \int \, dx \, dy \) into \( \int r \ dr \, d\theta\).

This Jacobian is fundamental in obtaining the correct expressions when transforming between different coordinate systems in multivariable integration.

Region of Integration

Understanding the region of integration is crucial when converting and evaluating integrals. The region of integration determines the limits of your integral and thus the bounds on \(r\) and \(\theta\) when using polar coordinates. In the original Cartesian integral, the limits form a semicircle in the upper half of the Cartesian plane, with a radius of 2.When converted to polar coordinates, this translates to:

• The region where \(0 \leq r \leq 2\), reflecting the radial component from the origin to the edge of the circle.
• \(0 \leq \theta \leq \frac{\pi}{2}\), specifying the angular sweep from the positive x-axis to the positive y-axis, forming a semicircle.

These bounds in polar coordinates reflect the geometric constraints of the problem accurately, and understanding these conversions helps ensure that the correct areas and values are calculated when performing the integration.