Question

Sketch the following regions \(R\). Then express \(\iint_{R} f(r, \theta) d A\) as an iterated integral over \(R.\) The region outside the circle \(r=1\) and inside the rose \(r=2 \sin 3 \theta\) in the first quadrant

Short answer

Question: Express the double iterated integral over the regions defined as the area outside the circle r = 1 and inside the rose r = 2sin3θ in the first quadrant as a definite integral. Sketch the regions and find the intersection points. Answer: The regions outside the circle r=1 and inside the rose r=2sin3θ can be expressed as a double iterated integral as follows: $$\iint_{R}f(r, \theta)dA = \int_{\frac{\pi}{9}}^{\frac{5\pi}{9}} \int_{1}^{2\sin 3\theta} f(r, \theta) r \ dr \ d\theta.$$ The intersection points between the circle and the rose are found at angles: $$\theta_1 = \frac{\pi}{9}, \quad \theta_2 = \frac{5\pi}{9}, \quad \theta_3 = \frac{3\pi}{9}.$$

Step-by-step solution

Sketch the regions

First, let's sketch the circle with polar equation \(r = 1\) and the rose with polar equation \(r = 2 \sin 3\theta\). The circle has a radius of one and is centered at the origin, while the rose has three petals since the coefficient in front of \(\theta\) is \(3\). The length of each petal is \(2\). Sketching both the circle and the rose on the polar grid, we can see the regions outside the circle and inside the rose.

Find intersection points

To express the given integral over the regions \(R\), we need to find the intersection points between the circle and the rose. This can be done by setting their polar equations equal to each other: $$1 = 2\sin 3\theta.$$ Divide by 2: $$\frac{1}{2} = \sin 3\theta.$$ Now, we need to find the values of \(\theta\) that satisfy this equation in the first quadrant. Inverse the sine function: $$\theta_{1,2,3} = \frac{1}{3}(\arcsin\frac{1}{2} + 2n\pi), \quad n = 0, 1, 2 .$$ Now we have the following \(\theta\) values that correspond to the intersection points: $$\theta_1 = \frac{\pi}{9}, \quad \theta_2 = \frac{5\pi}{9}, \quad \theta_3 = \frac{3\pi}{9}.$$

Setup the iterated integral

Now we have the limits of integration for \(\theta\). We know that the region is outside the circle \(r=1\) and inside the rose \(r=2\sin 3\theta\), therefore, our integral becomes: $$\iint_{R}f(r, \theta)dA = \int_{\frac{\pi}{9}}^{\frac{5\pi}{9}} \int_{1}^{2\sin 3\theta} f(r, \theta) r \ dr \ d\theta.$$ Now the double integral is expressed as an iterated integral over region \(R\).

Key concepts

Iterated Integrals

Iterated integrals are a fundamental part of multivariable calculus. They help us evaluate double integrals by breaking them down into a sequence of single integrals. In polar coordinates, an iterated integral takes the form:
\[\iint_{R} f(r, \theta) \, dA = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r, \theta) r \, dr \, d\theta.\]Here, we integrate first with respect to the radial component \(r\) and then the angular component \(\theta\). The additional \(r\) in the integrand accounts for the conversion from Cartesian to polar coordinates, as polar areas scale differently.To set up an iterated integral correctly, identify:

• The limits for \(r\), which might be a constant or a function of \(\theta\).
• The limits for \(\theta\), which usually span an interval based on the problem (often a sector of a circle).

Understanding these limits allows a seamless transition from a general double integral to the iterated form, unlocking possible ways to evaluate complex areas.

Region Sketching

Region sketching in polar coordinates can seem daunting, but it's a visual step that makes solving integrals much easier. The goal is to identify and illustrate the boundaries of the region over which we are integrating.For the given example:

• First, plot the circle \(r = 1\), which is simply a circle with radius 1 centered at the origin.
• Next, sketch the rose \(r = 2 \sin 3\theta\), noting its three petals because the coefficient of \(\theta\) is 3, and the petal length is 2.

You'll want to focus on the first quadrant since that’s the region of interest. By sketching both, you visualize where the region lies—outside the circle but inside the rose.
This provides a clearer understanding of where the boundaries are, which is essential for setting the limits of the iterated integral properly.

Intersection Points

Finding intersection points between different regions in polar coordinates is crucial for defining integration limits. This involves solving for points where the equations of the boundaries are equal.For our region, set:\[1 = 2 \sin 3\theta.\]Solving gives:\[\frac{1}{2} = \sin 3\theta.\]To find the angles that satisfy this equation, calculate:\[\theta = \frac{1}{3}(\arcsin\frac{1}{2} + 2n\pi),\]where \(n\) is an integer.In the first quadrant, these values are \(\theta_1 = \frac{\pi}{9}\), \(\theta_2 = \frac{5\pi}{9}\), and \(\theta_3 = \frac{3\pi}{9}\).
These points not only define where the boundaries of the circle and rose intersect but also set the limits for \(\theta\) in your integral setup. Accurately identifying intersection points is a step toward ensuring that the integration is over the correct region.