Evaluate The Double Integral R(3xy)dA, Where R Is The Region In The First Quadrant Enclosed By The Circle
Understanding how to evaluate double integrals over specific regions is fundamental in multivariable calculus. In particular, integrals involving regions enclosed by circles are common and serve as excellent examples for illustrating coordinate transformations and integration techniques. This article explores the detailed process of evaluating the double integral \( \iint_R 3xy \, dA \), where \( R \) is the region in the first quadrant enclosed by a circle.
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Introduction to Double Integrals and Their Significance
Double integrals extend the concept of single-variable integrals into two dimensions, allowing the calculation of areas, volumes, and other quantities over a specified region in the plane. They are crucial in fields such as physics, engineering, economics, and probability, where the quantity of interest depends on two variables.
The general form of a double integral over a region \( R \) is:
\[
\iint_R f(x, y) \, dA
\]
where \( f(x, y) \) is a function defined over \( R \). The region \( R \) can be described using various coordinate systems, most notably Cartesian or polar coordinates, depending on its shape.
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Understanding the Region \( R \)
Before computing the integral, it's essential to understand the region \( R \). Given that:
- \( R \) is in the first quadrant (where \( x \geq 0 \), \( y \geq 0 \))
- \( R \) is enclosed by a circle
Typically, the circle in question is of the form:
\[
x^2 + y^2 = r^2
\]
where \( r \) is the radius of the circle.
Assumption: For the purpose of this evaluation, we assume \( R \) is the region in the first quadrant enclosed by the circle \( x^2 + y^2 = a^2 \), with \( a > 0 \). This is a common and illustrative case.
Region \( R \):
\[
R = \left\{ (x, y) \mid x \geq 0, \, y \geq 0, \, x^2 + y^2 \leq a^2 \right\}
\]
Visually, \( R \) is a quarter circle in the first quadrant with radius \( a \).
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Choosing the Appropriate Coordinate System
Given the circular boundary, converting to polar coordinates simplifies the evaluation.
Polar coordinates are defined as:
\[
x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta
\]
where:
- \( r \geq 0 \)
- \( \theta \) is the angle measured from the positive \( x \)-axis (usually in radians).
Region \( R \) in polar coordinates:
- \( r \) ranges from 0 to \( a \)
- \( \theta \) ranges from 0 to \( \pi/2 \) (since the region is in the first quadrant)
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Setting Up the Double Integral in Polar Coordinates
Transforming the integrand:
\[
f(x, y) = 3xy
\]
becomes:
\[
3xy = 3 \times (r \cos \theta) \times (r \sin \theta) = 3 r^2 \cos \theta \sin \theta
\]
The differential area element \( dA \):
\[
dA = r \, dr \, d\theta
\]
Thus, the double integral over \( R \):
\[
\iintR 3xy \, dA = \int{\theta=0}^{\pi/2} \int_{r=0}^a 3 r^2 \cos \theta \sin \theta \times r \, dr \, d\theta
\]
Simplifies to:
\[
\int{0}^{\pi/2} \int{0}^{a} 3 r^3 \cos \theta \sin \theta \, dr \, d\theta
\]
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Evaluating the Integral Step-by-Step
Step 1: Integrate with respect to \( r \)
\[
\int{0}^{a} 3 r^3 \, dr = 3 \times \frac{r^4}{4} \Big|{0}^{a} = \frac{3}{4} a^4
\]
Step 2: Integrate with respect to \( \theta \)
Now, the integral reduces to:
\[
\frac{3}{4} a^4 \int_{0}^{\pi/2} \cos \theta \sin \theta \, d\theta
\]
Recall the identity:
\[
\sin 2\theta = 2 \sin \theta \cos \theta
\]
which implies:
\[
\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta
\]
Thus:
\[
\int{0}^{\pi/2} \cos \theta \sin \theta \, d\theta = \frac{1}{2} \int{0}^{\pi/2} \sin 2\theta \, d\theta
\]
Calculate this integral:
\[
\frac{1}{2} \times \left[ - \frac{1}{2} \cos 2\theta \right]_{0}^{\pi/2} = - \frac{1}{4} \left[ \cos \pi - \cos 0 \right]
\]
Evaluate:
\[
\cos \pi = -1, \quad \cos 0 = 1
\]
So:
\[
- \frac{1}{4} (-1 - 1) = - \frac{1}{4} (-2) = \frac{1}{2}
Step 3: Final calculation
Multiply the results:
\[
\frac{3}{4} a^4 \times \frac{1}{2} = \frac{3}{8} a^4
\]
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Final Result of the Double Integral
\[
\boxed{
\iint_R 3xy \, dA = \frac{3}{8} a^4
}
\]
This elegantly compact result illustrates how coordinate transformations and identities simplify what initially looks like a complex integral.
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Additional Considerations and Variations
While the above example assumes a specific circle \( x^2 + y^2 = a^2 \), similar procedures apply to other circular regions or different radii. Adjusting the bounds accordingly allows for flexible applications.
Other possible regions:
- Complete circle (both quadrants): Would require integrating over \( \theta \in [0, 2\pi] \)
- Different circle equations: For example, \( (x - h)^2 + (y - k)^2 = r^2 \), with shifts in the circle's center, may necessitate coordinate translation or different methods.
Using Cartesian coordinates:
In some cases, Cartesian bounds are more straightforward if the region is a segment between two curves. But for circular boundaries, polar coordinates generally simplify the process.
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Applications of the Evaluation
Understanding this integral evaluation has practical applications:
- Physics: Calculating mass or charge distribution over circular regions.
- Engineering: Computing heat flow or material properties in circular domains.
- Probability: Finding expected values where the joint distribution is defined over a circular region.
- Mathematics: Demonstrating techniques of variable substitution, symmetry, and integral identities.
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Conclusion
Evaluating double integrals over regions enclosed by circles in the first quadrant involves recognizing the symmetry and simplicity offered by polar coordinates. Converting integrals into polar coordinates reduces the complexity of boundary conditions and simplifies the integrand. The integral \( \iint_R 3xy \, dA \) over a quarter circle of radius \( a \) evaluates neatly to \( \frac{3}{8} a^4 \), demonstrating the power of coordinate transformations and integral identities.
Mastering these techniques enhances problem-solving skills in advanced calculus and provides foundational tools for tackling a wide range of problems involving circular regions in two dimensions. Whether dealing with theoretical mathematics or applied sciences, understanding how to evaluate such integrals is an essential skill in the mathematician's toolkit.
