Problem 16 - Thomas Calculus

Question

In Exercises \(9-22,\) change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$ \int_{\sqrt{2}}^{2} \int_{\sqrt{4-y^{2}}}^{y} d x d y $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{4} \).

Step 1: Identify the Cartesian Region of Integration

For the given Cartesian integral, we have the limits for \(y\) as \( \sqrt{2} \leq y \leq 2 \), and for a fixed \(y\), the limits for \(x\) are \( \sqrt{4-y^2} \leq x \leq y \). Visualize this region in the Cartesian plane.

Step 2: Plot the Cartesian Bounds

The curves \(x = \sqrt{4 - y^2}\) and \(x = y\) are boundaries of the region in the xy-plane. The curve \(x = \sqrt{4 - y^2}\) represents a semicircle of radius 2 centered at the origin on the positive x-side. The line \(x = y\) is a diagonal line, and the limits \(y = \sqrt{2}\) to \(y = 2\) indicate a specific segment of that region.

Step 3: Convert to Polar Coordinates

To convert to polar coordinates, use the transformations \(x = r \cos \theta\) and \(y = r \sin \theta\). Also, note that the Jacobian of the transformation is \(r\). The relationship between \(x\), \(y\), and \(r\) is \(r^2 = x^2 + y^2\) and allows for easier substitution.

Step 4: Determine the Polar Bounds for \(r\) and \(\theta\)

The semicircle \(x = \sqrt{4 - y^2}\) corresponds to \(r = 2\), and the line \(x = y\) corresponds to \(\theta = \frac{\pi}{4}\). The range for \(r\) is from \(\sqrt{2}\) to \(2\) and for \(\theta\) from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\).

Step 5: Set Up the Polar Integral

The integral converts to \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{\sqrt{2}}^{2} r \, dr \, d\theta \). The \(r\) in the integral represents the Jacobian determinant from the change of coordinates.

Step 6: Evaluate the Polar Integral

Evaluate the integral, starting with the inner integral with respect to \(r\): \[ \int_{\sqrt{2}}^{2} r \, dr = \left[ \frac{r^2}{2} \right]_{\sqrt{2}}^{2} = \left( \frac{4}{2} - \frac{2}{2} \right) = 1. \] Now, evaluate the outer integral: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \, d\theta = \left[ \theta \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \]

Step 7: Conclusion

The value of the original Cartesian integral, converted and evaluated in polar coordinates, is \( \frac{\pi}{4} \).

Key Concepts

Cartesian Coordinates

Cartesian coordinates, named after the French mathematician René Descartes, are the most common way to locate points in a plane using a pair of numerical values. These values, referred to as the x-coordinate and y-coordinate, describe the point's position concerning two perpendicular directed lines, called axes.

• The x-axis runs horizontally, while the y-axis extends vertically.
• The point where these axes intersect is known as the origin, denoted as (0,0).

In the given exercise, we deal with a Cartesian integral defined over specific limits for both the x and y-coordinates:

• The range for y is from \( \sqrt{2} \) to 2.
• For each y, the range for x is bounded by two curves: \( x = \sqrt{4-y^2} \) and \( x = y \).

These boundaries form a region in the Cartesian plane whose area we can calculate by converting to polar coordinates.

Integral Conversion

Integral conversion involves changing the form of a given integral, often to simplify computation. In this context, we are converting a Cartesian integral into a polar integral.

• Polar coordinates consist of a pair: \( (r, \theta) \), where \( r \) is the radial distance from the origin and \( \theta \) is the angular position.
• The transformation equations are \( x = r \cos\theta \) and \( y = r \sin\theta \).

Integrals in polar coordinates utilize these transformations and often reveal simpler bounds or symmetries in a given problem. In our exercise:- The transformation allows us to see that the circular segment and diagonal line become straightforward bounds in polar form.

Region of Integration

The region of integration is a crucial step in integrating a function over a specified area. In Cartesian coordinates, this involves visualizing the area determined by the given limits of x and y, as previously defined.Once you identify these bounds, switching to polar coordinates can often simplify the problem. Polar coordinates are particularly useful when the region of integration is circular or has radial symmetry. In our problem:

• The semicircle boundary \( x = \sqrt{4-y^2} \) turns into \( r = 2 \), demonstrating a constant radius wall at \(r=2\).
• The line \( x = y \) forms a boundary at \( \theta = \frac{\pi}{4} \), with the range of integration going up to \( \theta = \frac{\pi}{2} \).

Jacobian Determinant

The Jacobian determinant is a critical concept in the transformation of variables in multiple integrals when converting between coordinate systems. It accounts for area scaling during transformation.In our case, as we switch from Cartesian to polar coordinates, the Jacobian determinant is \( r \). This term is derived from the partial derivatives of the transformation equations and effectively adjusts for the 'squishing' or 'stretching' of space due to the coordinate change.Here's what happens:

• When converting the original integral to polar coordinates, we multiply by \( r \) to account for this area change.
• This factor ensures the integral computes the correct area under the new system.

Thus, the Jacobian determinant helps keep the areas under the curves equivalent through transformation, ensuring the same physical region is integrated in both systems.