Question

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

Short answer

Based on the given step-by-step solution, the iterated integral setup that covers the region R, which is inside the leaf of the rose r = 2sin(2θ) in the first quadrant, is: $$ \iint_{R} f(r, \theta) d A = \int_{0}^{\frac{\pi}{4}} \int_{0}^{2 \sin(2 \theta)} f(r, \theta) r \, dr \, d\theta. $$

Step-by-step solution

Determine the boundaries of the first quadrant leaf

Before drawing the rose, we need to find the θ values where the rose intersects the axes of the polar coordinate system. To do that, we set r=0 and solve for θ: $$ 0 = 2 \sin(2\theta) $$ Now solve for θ values in the first quadrant (0 ≤ θ ≤ π/2): $$ \theta_1 = 0 \\ \theta_2 = \frac{\pi}{4} $$

Sketch the first quadrant leaf

Using the information from Step 1, we know that the leaf in the first quadrant starts at the origin (0,0) and ends at the top-right part of the rose when θ = π/4. Start by drawing a polar coordinate system, then draw the curve r = 2sin(2θ) from θ = 0 to θ = π/4. This will create a petal-like form that resides in the first quadrant of the polar coordinate system.

Set up the iterated integral

Now that we have the sketch of region R, we can set up the iterated integral to cover this region in the polar coordinate system. In polar coordinates, the double integral of a function, f(r,θ), can be expressed as: $$ \iint_R f(r, \theta) dA = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r, \theta) r \, dr \, d\theta $$ From Step 1, we know that the range of θ for the first quadrant leaf of region R is 0 to π/4 (θ_1 = 0, θ2 = π/4). For r, the range is from 0 to the curve defined by the function r = 2sin(2θ) (r_1 = 0, r_2 = 2sin(2θ)). Plug these boundaries into the iterated integral to obtain: $$ \iint_{R} f(r, \theta) d A = \int_{0}^{\frac{\pi}{4}} \int_{0}^{2 \sin(2 \theta)} f(r, \theta) r \, dr \, d\theta $$

Key concepts

Iterated Integrals

In calculus, iterated integrals are a way to compute the volume under a surface by breaking it down into smaller parts. Imagine peeling an orange slice by slice. Each slice can be thought of as a smaller integral.
When working in polar coordinates, iterated integrals let you evaluate double integrals more easily by integrating over one variable at a time. You first integrate with respect to one variable while treating the other variable as a constant.

• Here, we have a function depending on two variables, usually represented as \(f(r, \theta)\).
• The integral is performed first over \(r\) for a fixed \(\theta\), then over \(\theta\).
• This allows for managing complex shapes by iterating through each variable.

For example, in the exercise, after sketching the rose curve and identifying its bounds, an iterated integral was set up to cover the first quadrant of the polar coordinate system.
This involves integrating \(f(r, \theta)r\) first from \(r_1\) to \(r_2\), and then from \(\theta_1\) to \(\theta_2\). It ensures the correct region is covered in the integration process.

Polar Coordinate System

The polar coordinate system is an alternative to the Cartesian coordinate system, often useful for dealing with circular or spiral patterns. Instead of using perpendicular axes, polar coordinates use a center point and measure distances and angles.
In polar coordinates:

• \(r\) represents the distance from the origin to the point.
• \(\theta\) is the angle from the positive x-axis to the line connecting the origin and the point.

Considering the exercise, a polar curve is given by \(r=2 \sin 2 \theta\), a form known as a *rose curve*. These curves can produce beautiful patterns of petals, making them ideal for using the polar grid.
To express regions in polar coordinates, it's important to carefully identify the ranges for \(r\) and \(\theta\). For the rose curve in the first quadrant, \(\theta\) ranges from 0 to \(\frac{\pi}{4}\) which defines part of one petal.
This method is powerful in converting more complex integration problems into simpler ones by changing the coordinate system itself.

Double Integrals

Double integrals extend the concept of integration to two dimensions, allowing you to calculate areas and volumes under surfaces. They are central when it comes to finding area in a plane region or the volume under a surface.
In a Cartesian plane, double integrals use the form \(\int \int f(x, y) \, dx \, dy\).
However, converting to polar coordinates transforms the integral into \(\iint f(r, \theta) r \, dr \, d\theta\), due to the way area elements are defined in this coordinate system.

• The inner integral integrates with respect to \(r\), from the origin outward.
• The outer integral covers the full angular range for \(\theta\).
• The additional \(r\) in the integral accounts for the differing sizes of the circles as you move out from the center.

In the rose curve exercise example, the double integral calculates the area under the surface defined over the petal shape. This requires attention to both the shape of the region _as well_ as the function describing the surface.