Chapter 13: Problem 6
In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. \(f(x, y)=4 ; R\) is the region enclosed by the petal of the rose curve \(r=\sin (2 \theta)\) in the first quadrant.
Short Answer
Expert verified
The integral evaluates to \( \frac{\pi}{2} \).
Step by step solution
01
Understand the Problem
We need to compute the double integral \( \iint_{R} f(x, y) \, dA \) over a given region \( R \), where \( f(x, y) = 4 \). The region \( R \) is the area enclosed by the petal of the rose curve \( r = \sin(2\theta) \) in the first quadrant.
02
Recognize the Region in Polar Coordinates
The rose curve \( r = \sin(2\theta) \) forms a petal in the first quadrant between \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \). The radial component \( r \), in this case, varies from 0 to \( \sin(2\theta) \) for a given \( \theta \).
03
Set up the Integral in Polar Coordinates
To express the integral in polar coordinates, use \( x = r\cos\theta \) and \( y = r\sin\theta \). The differential area element \( dA \) becomes \( r \, dr \, d\theta \). Thus, the double integral is set up as: \[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sin(2\theta)} 4r \, dr \, d\theta \]
04
Evaluate the Inner Integral
First, integrate with respect to \( r \): \[ \int_{0}^{\sin(2\theta)} 4r \, dr = \left[ 2r^2 \right]_{0}^{\sin(2\theta)} = 2(\sin(2\theta))^2 \]
05
Simplify the Expression
Recall that \( (\sin(2\theta))^2 = \frac{1 - \cos(4\theta)}{2} \), so the result from Step 4 becomes: \[ 2 \cdot \frac{1 - \cos(4\theta)}{2} = 1 - \cos(4\theta) \]
06
Evaluate the Outer Integral
Now, integrate with respect to \( \theta \): \[ \int_{0}^{\frac{\pi}{2}} (1 - \cos(4\theta)) \, d\theta \] Breaking this into simpler parts: \[ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \int_{0}^{\frac{\pi}{2}} \cos(4\theta) \, d\theta \]
07
Compute Each Part
For \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta \), the result is simply \( \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \).For \( \int_{0}^{\frac{\pi}{2}} \cos(4\theta) \, d\theta \), use the substitution \( u = 4\theta \), which gives: \[ \frac{1}{4} \int_{0}^{2\pi} \cos(u) \, du = \frac{1}{4} [\sin(u)]_{0}^{2\pi} = 0 \]
08
Compute the Final Result
Combine the results from Step 7: \[ \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Therefore, the value of the double integral is \( \frac{\pi}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integral
A double integral is a powerful tool in calculus that allows us to compute volumes under surfaces or find areas of more complex shapes. This concept extends the idea of a single integral, which deals only with functions of a single variable, to functions of two variables, say \( f(x, y) \). The double integral of a function over a region \( R \) is represented as \( \iint_R f(x, y) \, dA \).
Here's a breakdown of what this means:
Here's a breakdown of what this means:
- \( f(x, y) \) is the function you are integrating.
- \( R \) is the specific region in the plane over which you are integrating.
- \( dA \) denotes a small element of area in the plane.
Polar Coordinates
Polar coordinates are a system of mathematical coordinates that works well for integrating regions with circular symmetry, like circles or petals. In this system, points are determined by a radius \( r \) and an angle \( \theta \) instead of \( x \) and \( y \, \) as in Cartesian coordinates.
When using polar coordinates, you convert the original \( x \) and \( y \) values using the formulas:
When using polar coordinates, you convert the original \( x \) and \( y \) values using the formulas:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
Rose Curve
A rose curve is a fascinating graph that looks like petals forming a flower. It's described using polar coordinates by the equation \( r = \sin(n\theta) \) or \( r = \cos(n\theta) \), where \( n \) determines the number of petals.
Key insights for rose curves:
Key insights for rose curves:
- If \( n \) is even, the rose curve will have \( 2n \) petals.
- If \( n \) is odd, the rose curve will have \( n \) petals.
Trigonometric Substitution
Trigonometric substitution is a useful technique for simplifying integrals, especially when the integrand involves complex trigonometric expressions. This technique exploits trigonometric identities to transform the integrand into a simpler form.
For example:
For example:
- We used the identity \( (\sin(2\theta))^2 = \frac{1 - \cos(4\theta)}{2} \) to simplify expressions involving \( (\sin(2\theta))^2 \).