Evaluate The Iterated Integral By Converting To Polar Coordinates. ./2 - Y2 5(x + Y) Dx Dy 12- 2v2 3
When it comes to evaluating complex double integrals, especially those involving non-trivial regions or integrand functions, converting to polar coordinates often simplifies the process significantly. This approach is particularly useful when the region of integration is circular, elliptical, or can be more easily described in terms of radius and angle rather than Cartesian coordinates. Today, we will explore how to evaluate the iterated integral given by:
\[
\iint_{R} ( \text{expression} ) \, dx\,dy
\]
where the integral's bounds and the integrand suggest a transformation to polar coordinates, making the evaluation more straightforward. The specific integral in question appears to involve terms such as \(Y^2\), \(x + Y\), and constants like 12 and \(2\sqrt{3}\). Although the original notation appears somewhat fragmented, we will interpret and reconstruct the integral carefully to guide you through the process.
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Understanding the Integral and Its Region of Integration
Before converting to polar coordinates, it’s crucial to interpret the integral's bounds and integrand correctly.
1. Deciphering the Integral Expression
The notation appears to be:
\[
\iint_{D} \left( 2 - y^2 + 5(x + y) \right) \, dx\,dy
\]
or similar, with bounds related to 12 and \(2\sqrt{3}\). Since the given expression is somewhat ambiguous, we'll assume the integral involves the function:
\[
f(x, y) = 2 - y^2 + 5(x + y)
\]
over a region \(D\). The bounds possibly relate to the region's limits, which might be a circle or a sector involving radii or angles corresponding to these constants.
2. Determining the Region \(D\)
Given the constants, a common region associated with such integrals involves circles or sectors:
- The constant 12 could relate to a radius squared, i.e., \(r^2 = 12\), implying \(r = \sqrt{12} = 2\sqrt{3}\).
- The mention of \(2\sqrt{3}\) supports this, suggesting the region might be a circle with radius \(2\sqrt{3}\).
Assuming the region \(D\) is a circle centered at the origin with radius \(2\sqrt{3}\):
\[
D: \quad x^2 + y^2 \leq 12
\]
which is a common scenario for converting to polar coordinates.
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Converting Cartesian Coordinates to Polar Coordinates
Conversion to polar coordinates is a standard technique when dealing with circular regions, as it simplifies the bounds and often the integrand.
1. Polar Coordinate Definitions
The transformation from Cartesian \((x, y)\) to polar \((r, \theta)\) is given by:
\[
x = r \cos \theta
\]
\[
y = r \sin \theta
\]
where:
- \(r \geq 0\) is the distance from the origin.
- \(\theta \in [0, 2\pi)\) is the angle measured from the positive \(x\)-axis.
2. Jacobian for the Transformation
The differential area element transforms as:
\[
dx\,dy = r\,dr\,d\theta
\]
This Jacobian determinant accounts for the change of variables from Cartesian to polar coordinates.
3. Expressing the Integrand in Polar Coordinates
Given the integrand:
\[
f(x, y) = 2 - y^2 + 5(x + y)
\]
substituting \(x = r \cos \theta\) and \(y = r \sin \theta\):
\[
f(r, \theta) = 2 - (r \sin \theta)^2 + 5(r \cos \theta + r \sin \theta)
\]
which simplifies to:
\[
f(r, \theta) = 2 - r^2 \sin^2 \theta + 5 r (\cos \theta + \sin \theta)
\]
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Setting Up the Integral in Polar Coordinates
Having established the transformation, the integral becomes:
\[
\iint{D} f(x, y) \, dx\,dy = \int{\theta=0}^{2\pi} \int_{r=0}^{R} f(r, \theta) \, r \, dr\, d\theta
\]
where \(R = 2\sqrt{3}\).
1. Final Form of the Integral
\[
\boxed{
\int{0}^{2\pi} \int{0}^{2 \sqrt{3}} \left[ 2 - r^2 \sin^2 \theta + 5 r (\cos \theta + \sin \theta) \right] r \, dr\, d\theta
}
\]
This integral can now be tackled by integrating with respect to \(r\) first, then \(\theta\).
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Evaluating the Integral Step-by-Step
Let's proceed systematically to evaluate the integral.
1. Expand the Integrand
The integrand:
\[
f(r, \theta) \times r = \left[ 2 - r^2 \sin^2 \theta + 5 r (\cos \theta + \sin \theta) \right] r
\]
becomes:
\[
I(r, \theta) = 2r - r^3 \sin^2 \theta + 5 r^2 (\cos \theta + \sin \theta)
\]
2. Integrate with Respect to \(r\)
The limits are from \(0\) to \(2\sqrt{3}\):
\[
\int_{0}^{2\sqrt{3}} \left( 2r - r^3 \sin^2 \theta + 5 r^2 (\cos \theta + \sin \theta) \right) dr
\]
Break this into three separate integrals:
\[
I1(\theta) = \int{0}^{2\sqrt{3}} 2r\, dr
\]
\[
I2(\theta) = - \sin^2 \theta \int{0}^{2\sqrt{3}} r^3\, dr
\]
\[
I3(\theta) = 5 (\cos \theta + \sin \theta) \int{0}^{2\sqrt{3}} r^2\, dr
\]
Calculating each:
- \(I_1(\theta)\):
\[
\int{0}^{a} 2r\, dr = r^2 \bigg|{0}^{a} = a^2
\]
with \(a = 2 \sqrt{3}\):
\[
a^2 = (2 \sqrt{3})^2 = 4 \times 3 = 12
\]
So,
\[
I_1(\theta) = 12
\]
- \(I_2(\theta)\):
\[
- \sin^2 \theta \times \frac{r^4}{4} \bigg|_{0}^{2 \sqrt{3}} = - \sin^2 \theta \times \frac{(2 \sqrt{3})^4}{4}
Calculate \( (2 \sqrt{3})^4 \):
\[
(2 \sqrt{3})^4 = ( (2)^4 ) \times (\sqrt{3})^4 = 16 \times (3^2) = 16 \times 9 = 144
\]
Thus,
\[
I_2(\theta) = - \sin^2 \theta \times \frac{144}{4} = - \sin^2 \theta \times 36
\]
- \(I_3(\theta)\):
\[
5 (\cos \theta + \sin \theta) \times \frac{r^3}{3} \bigg|_{0}^{2 \sqrt{3}} = 5 (\cos \theta + \sin \theta) \times \frac{(2 \sqrt{3})^3}{3}
\]
Calculate \( (2 \sqrt{3})^3 \):
\[
(2 \sqrt{3})^3 = 2^3 \times (\sqrt{3})^3 = 8 \times 3 \sqrt{3} = 8 \times 3 \sqrt{3} = 24 \sqrt{3}
\]
Therefore:
\[
I_3(\theta) = 5 (\cos \theta + \sin \theta) \times \frac{24 \sqrt{3}}{3} = 5 (\cos \theta + \sin \theta) \times 8 \sqrt{3}
\]
Simplify:
\[
I_3(\theta) = 40 \sqrt{3} (\cos \theta + \sin \theta)
\]
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3. Summing the Results
The total integral over \(r\) for a fixed \(\theta\):
\[
J(\
