Evaluate The Iterated Integral By Converting To Polar Coordinates

Evaluating the Iterated Integral by Converting to Polar Coordinates

When a double integral over a region in the Cartesian plane looks messy or impossible to solve directly, a classic trick is to switch to polar coordinates. This transformation often simplifies the integrand and the limits of integration, turning a daunting iterated integral into a manageable one. Below we walk through the theory, the steps involved, and a complete worked example that illustrates why polar coordinates can be a game‑changer for many integrals.

Introduction

An iterated integral is a double integral written as a nested single integral, for instance
[ \int_{a}^{b}!!But \int_{c(x)}^{d(x)} f(x,y),dy,dx. ] If the region bounded by (c(x)) and (d(x)) is a circle, an annulus, or any shape that is naturally described in terms of radius and angle, the Cartesian form can be cumbersome. Now, by converting to polar coordinates ((r,\theta)), where
[ x = r\cos\theta,\qquad y = r\sin\theta, ] the area element transforms as (dx,dy = r,dr,d\theta). This extra factor of (r) often cancels parts of the integrand or simplifies the limits.

The key benefits of the polar transformation are:

1. Simplified limits: Circles and sectors become constant bounds in (r) and (\theta).
2. Symmetry exploitation: Radial symmetry in the integrand reduces complexity.
3. Avoiding awkward algebra: Square roots and quadratic expressions disappear in favor of linear terms in (r).

Below we outline the systematic approach to converting an iterated integral to polar coordinates and then apply it to a representative example.

Step‑by‑Step Conversion Process

1. Identify the Region (R)
   Sketch or describe the domain of integration. Determine whether it is a circle, annulus, sector, or a shape that can be expressed in polar form And that's really what it comes down to..

2. Express the Boundary in Polar Form
   Convert each Cartesian boundary equation into polar coordinates using (x = r\cos\theta) and (y = r\sin\theta).

   • For a circle (x^2 + y^2 = a^2), we get (r = a).
   • For a line (y = mx + b), we get (r\sin\theta = m(r\cos\theta)+b).

3. Determine the Limits for (r) and (\theta)

   • (\theta) limits come from the angular extent of the region.
   • (r) limits are usually functions of (\theta) (inner radius to outer radius).
     Ensure the limits are consistent and cover the entire region without overlap or omission.

4. Transform the Integrand
   Replace (x) and (y) in (f(x,y)) with (r\cos\theta) and (r\sin\theta).
   Multiply the resulting expression by the Jacobian determinant (r).

5. Set Up the Polar Integral
   The new iterated integral takes the form
   [ \int_{\theta_1}^{\theta_2}!!\int_{r_{\text{inner}}(\theta)}^{r_{\text{outer}}(\theta)} f(r\cos\theta,r\sin\theta), r,dr,d\theta. ]

6. Evaluate
   Perform the inner integral with respect to (r) first, then integrate the resulting expression with respect to (\theta). Use standard techniques: substitution, integration by parts, or known antiderivatives.

Worked Example

Problem
Evaluate the iterated integral
[ I = \int_{0}^{1}!!\int_{\sqrt{1-y^2}}^{1}!!x,dx,dy. ]

Interpretation
The inner integral limits (x = \sqrt{1-y^2}) to (x = 1) suggest a right half of a circle of radius 1, shifted to the right. The outer limits (y = 0) to (y = 1) confine us to the upper quarter of the circle. Thus, the region (R) is a quarter‑disk of radius 1 in the first quadrant, bounded on the left by the circle (x^2 + y^2 = 1) and on the right by the line (x=1).

1. Convert the Region to Polar Coordinates

• The circle (x^2 + y^2 = 1) becomes (r = 1).
• The line (x = 1) becomes (r\cos\theta = 1), i.e. (r = \sec\theta).
• Since (y \ge 0) and (x \ge 0), (\theta) ranges from (0) to (\pi/2).

Thus, for each (\theta) in ([0,\pi/2]), the radial coordinate (r) runs from the circle (r = 1) outward to the line (r = \sec\theta). Note that (\sec\theta) is defined for (\theta \in[0,\pi/2)) and tends to infinity as (\theta \to \pi/2). That said, the intersection of the line (x=1) with the circle occurs at (\theta = 0), so the upper bound of (\theta) is actually (\theta = \arcsin(1/2) = \pi/6) when the line meets the circle Simple, but easy to overlook..

Set (x=1) and (x^2+y^2=1). Therefore the angular range is from (\theta=0) to (\theta=\pi/2), but the radial upper bound is (\min(1,\sec\theta)). So the line (x=1) touches the circle only at the point ((1,0)). For (\theta) between (0) and (\arccos(1)) (which is 0), (\sec\theta > 1). So the radial upper bound is (\sec\theta) when (\sec\theta \le 1), but that never happens. Hence the region is simply the quarter disk (0\le r \le 1), (0\le \theta \le \pi/2). On top of that, substituting (x=1) gives (1 + y^2 = 1 \implies y=0). Here's the thing — the line (x=1) does not cut the disk interior; it lies outside except at the boundary. And actually, for (\theta) in ([0,\pi/2]), (\sec\theta \ge 1). Thus the region is actually bounded by the circle and the line (x=1) only along the segment from ((1,0)) to ((1,1)). Thus the integral simplifies to the quarter disk.

Conclusion: The region (R) is the quarter disk (0\le r \le 1), (0\le \theta \le \pi/2).

2. Transform the Integrand

The integrand (x) becomes (r\cos\theta). Including the Jacobian (r), the integrand in polar form is
[ x,dx,dy ;\longrightarrow; (r\cos\theta), (r,dr,d\theta) = r^2\cos\theta,dr,d\theta. ]

3. Set Up the Polar Integral

[ I = \int_{0}^{\pi/2}!!\int_{0}^{1}! r^2\cos\theta,dr,d\theta. ]

4. Evaluate

Inner integral (with respect to (r))
[ \int_{0}^{1} r^2,dr = \left[\frac{r^3}{3}\right]_{0}^{1} = \frac{1}{3}. ]

Outer integral (with respect to (\theta))
[ \int_{0}^{\pi/2} \cos\theta,d\theta = \left[\sin\theta\right]_{0}^{\pi/2} = 1. ]

Combine
[ I = \frac{1}{3} \times 1 = \boxed{\frac{1}{3}}. ]

Thus, by converting to polar coordinates, the integral collapses to a simple product of two elementary integrals Worth knowing..

Scientific Explanation

The transformation hinges on the Jacobian determinant of the polar change of variables. The mapping ((r,\theta) \mapsto (x,y) = (r\cos\theta, r\sin\theta)) stretches or compresses area elements by a factor of (r). Geometrically, as you move outward from the origin, concentric circles grow in circumference; the area of an infinitesimal annular sector of radius (r) and angle (d\theta) is (r,dr,d\theta). This factor ensures that the total area computed in polar coordinates matches that in Cartesian coordinates.

When the integrand contains powers of (x) and (y) that combine into (x^2+y^2), the switch to (r^2) eliminates the cross terms and often linearizes the expression. Symmetry about the origin or an axis is naturally captured by the angular variable, allowing us to integrate over a single angular interval rather than splitting the region.

Frequently Asked Questions

Conclusion

Converting an iterated integral to polar coordinates is a powerful strategy that turns a potentially complex Cartesian integral into a tidy, often separable problem. In real terms, by carefully mapping the region, adjusting the limits, and incorporating the Jacobian factor, you can exploit radial symmetry and simplify the integrand dramatically. Mastery of this technique not only saves time but also deepens your geometric intuition about area and volume in the plane. Whether tackling textbook exercises or real‑world applications, polar coordinates remain an essential tool in the mathematician’s toolkit It's one of those things that adds up. Still holds up..