Problem 11 - Thomas Calculus

Question

In Exercises \(9-22,\) 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 equivalent polar integral evaluates to \(2\pi\).

Step 1: Identify the Region of Integration

The given integral is \[\int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}}(x^{2}+y^{2}) \, dx \, dy\].The limits on \(y\) are from 0 to 2, and for \(x\), they are from 0 to \(\sqrt{4-y^2}\). The upper limit for \(x\) is the equation of a semicircle, implying the region of integration describes a quarter circle in the first quadrant with radius 2.

Step 2: Convert to Polar Coordinates

In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). The expression \(x^2 + y^2\) becomes \(r^2\). The differential \(dx\,dy\) is replaced by \(r\,dr\,d\theta\). The region is a quarter circle, so \(r\) ranges from 0 to 2, and \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\). Thus, the integral becomes\[\int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \, dr \, d\theta\].

Step 3: Evaluate the Inner Integral

First, evaluate the inner integral with respect to \(r\):\[\int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{2} = \frac{16}{4} - 0 = 4\].

Step 4: Evaluate the Outer Integral

Now, evaluate the outer integral with respect to \(\theta\):\[\int_{0}^{\frac{\pi}{2}} 4 \, d\theta = 4\theta \bigg|_{0}^{\frac{\pi}{2}} = 4 \cdot \frac{\pi}{2} - 0 = 2\pi\].

Step 5: Final Outcome

The value of the polar integral, which is equivalent to the original Cartesian integral, is \(2\pi\).

Key Concepts

Cartesian integral

A Cartesian integral is a method used to calculate the area under a curve or the volume under a surface in the Cartesian coordinate system. In the given problem, the integral \( \int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}}(x^{2}+y^{2}) \, dx \, dy \) is an example of a Cartesian integral. Here, we compute the double integral over a region defined by the limits of integration for \( x \) and \( y \). The idea is to recognize the geometric region that these limits describe.

• The limits of integration for \( y \) are from 0 to 2.
• For \( x \), they are from 0 to \( \sqrt{4-y^2} \), which is the upper semi-circle.

The expression \( x^2 + y^2 \) in the integral hints at a form that may simplify when switching to another coordinate system.
Switching to polar coordinates often makes the integration easier, especially when dealing with circular or radial regions.

polar integral

A polar integral is an integration method used in polar coordinates, where the position of points is determined by a distance from a reference point and an angle from a reference direction. When converting a Cartesian integral to a polar integral, the functions, limits, and differentials must be adjusted accordingly.

• In polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \).
• The term \( x^2 + y^2 \) converts to \( r^2 \), simplifying some expressions.
• The differential \( dx \, dy \) is expressed as \( r \, dr \, d\theta \) in polar coordinates.

For our problem, the given Cartesian integral transforms into \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \, dr \, d\theta \), which is simpler to evaluate.
Polar integrals are particularly useful when dealing with circular or angular regions, as they align more naturally with the geometry of the problem.

quarter circle region

In this exercise, the region of integration is a quarter circle located in the first quadrant of the Cartesian plane. Understanding the region described by the integral is crucial for setting the correct limits when switching to polar coordinates. Here’s how to recognize and describe this region:

• The region is bounded by the semi-circle \( x^2 + y^2 = 4 \) and the lines \( y = 0 \) and \( x = 0 \).
• This means it is confined to the first quadrant, where both \( x \) and \( y \) are non-negative.
• In polar coordinates, this region is defined by \( r \) ranging from 0 to 2 and \( \theta \) from 0 to \( \frac{\pi}{2} \).

The identification of such regions simplifies the integration process because each region directly translates to simple constant limits in polar coordinates.
Recognizing it as a quarter circle is key as it influences the conversion of limits in the polar integral.

integration in polar coordinates

Integration in polar coordinates involves converting a function expressed in Cartesian coordinates into polar coordinates before evaluating the integral. This process is especially effective in regions with circular symmetry.

• The transformation involves substituting \( x = r\cos\theta \) and \( y = r\sin\theta \) into the function.
• The differential \( dx \, dy \) becomes \( r \, dr \, d\theta \), incorporating the Jacobian of the transformation.
• The range of \( r \) and \( \theta \) needs to be determined based on the region of integration.

In our problem, the conversion led to the simpler integral \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \, dr \, d\theta \).
Evaluating this integral starts with solving the inner integral with respect to \( r \), yielding 4, and then the outer integral with respect to \( \theta \), resulting in the final answer of \( 2\pi \). Integration in polar coordinates is a powerful technique that, when applied correctly, simplifies the evaluation of complex integrals.