Problem 10 - Thomas Calculus

Question

In Exercises \(9-22,\) change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right) d x d y $$

Step-by-Step Solution

Answer

The value of the polar integral is \( \frac{\pi}{8} \).

Step 1: Identify Region of Integration

The original Cartesian integral is given by two limits: \( x \) ranges from \( 0 \) to \( \sqrt{1-y^2} \), and \( y \) ranges from \( 0 \) to \( 1 \). This describes a region that is the first quadrant of a unit circle, specifically a quarter-circle from \( y \) axis to the \( x \) axis.

Step 2: Convert Cartesian to Polar Coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). Also, \( x^2 + y^2 = r^2 \). The Jacobian of the transformation from Cartesian to polar coordinates is \( r \). Thus, the integrand \( x^2 + y^2 \) becomes \( r^2 \), and the differential \( dx \ dy \) becomes \( r \, dr \, d\theta \).

Step 3: Set Up Polar Integral

The limits of integration change to reflect the geometry of the region. In terms of \( r \): from the origin to the unit circle, thus \( 0 \leq r \leq 1 \). For \( \theta \), the angle sweeps from \( 0 \) to \( \frac{\pi}{2} \) (quarter-circle). The new integral becomes \[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} r^3 \, dr \, d\theta. \]

Step 4: Evaluate the Inner Integral

Integrate \( r^3 \) with respect to \( r \): \[ \int_{0}^{1} r^3 \, dr = \left. \frac{r^4}{4} \right|_0^1 = \frac{1}{4}. \]

Step 5: Evaluate the Outer Integral

Substitute the evaluated inner integral result:\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{4} \, d\theta = \frac{1}{4} \left( \theta \right|_0^{\frac{\pi}{2})} = \frac{1}{4} \times \frac{\pi}{2} = \frac{\pi}{8}. \]

Step 6: State Final Answer

Thus, the value of the original Cartesian integral, expressed and evaluated in polar coordinates, is \( \frac{\pi}{8} \).

Key Concepts

Cartesian Integral

The Cartesian integral provided is often represented in the form \( \int \int f(x, y) \, dx \, dy \), which involves integrating over two variables, \( x \) and \( y \). In this exercise, the integral is set within particular bounds, where \( x \) varies from \( 0 \) to \( \sqrt{1-y^2} \), and \( y \) from \( 0 \) to \( 1 \). This setup defines a region that is the upper right quarter of a unit circle.

This integral calculates the area under the surface described by the integrand \( x^2 + y^2 \) over the given region. The complexity arises from the irregular bounds which represent a circular region. Thus, the Cartesian integral is converted into a polar integral, where the calculations are simpler given the circular symmetry of the region.

Unit Circle

The concept of the unit circle is crucial when dealing with polar coordinates. A unit circle is a circle with a radius of one, centered at the origin of the coordinate plane. In this problem, the region of integration corresponds to the quarter-unit circle located in the first quadrant.

In Cartesian coordinates, this is depicted by the area bounded by \( y \) values from \( 0 \) to \( 1 \) and \( x \) values from \( 0 \) to \( \sqrt{1-y^2} \), effectively cropping the typical unit circle from the \( x \)-axis up to the maximum point on the \( y \)-axis. This is why converting to polar coordinates simplifies the problem, as it perfectly matches the circular symmetry of the unit circle.

Transformation

A transformation from Cartesian to polar coordinates is utilized in this exercise to simplify the integration process over a circular region. In polar coordinates, a point \((x, y)\) is expressed as \((r, \theta)\), where \( r \) is the radial distance from the origin, and \( \theta \) is the angle from the positive \( x \)-axis.

• The relationships are \( x = r \cos \theta \) and \( y = r \sin \theta \).
• The integrand \( x^2 + y^2 \) transforms to \( r^2 \), as the equation of a circle \( x^2 + y^2 = r^2 \) holds.

By converting Cartesian limits to polar, we recognize the region as a section of a circle, which is best described by \( r \) from 0 to 1 and \( \theta \) from 0 to \( \frac{\pi}{2} \), depicting a quarter-circle. This transformation simplifies integration significantly.

Jacobian

The Jacobian is an essential factor when changing variables in multiple integrals, specifically when converting from Cartesian to polar coordinates. It accounts for the change of area element from \( dx \, dy \) to \( dr \, d\theta \).

• The Jacobian in polar coordinates is \( r \), which adjusts the area element since polar coordinates spread areas radially and angularly.
• This results in transforming \( dx \, dy \) into \( r \, dr \, d\theta \).

By incorporating the Jacobian \( r \) into the integral, it ensures that the change from rectangular to circular elements reflects the actual geometry of the region. Thus, the integrand \( r^2 \) is integrated with respect to \( r \, dr \, d\theta \) to accurately cover the entire quarter-region.