Problem 25 - Calculus Early Transcendentals

Question

Sketch the given region of integration \(R\) and evaluate the integral over \(R\) using polar coordinates. $$\iint_{R} 2 x y d A ; R=\left\\{(x, y): x^{2}+y^{2} \leq 9, y \geq 0\right\\}$$

Step-by-Step Solution

Answer

Answer: The value of the integral \(\iint_{R} 2xy dA\) over the region R is \(\frac{81}{4}\).

Step 1: Sketch the region R

The region R is described as \(\left\\{(x, y): x^{2}+y^{2} \leq 9, y \geq 0\right\\}\). This indicates that R is confined within a semicircle of radius 3, centered at the origin (0, 0) on a Cartesian plane. As \(y \geq 0\), we are only considering the upper half plane, making this a semicircle bounding the region above the x-axis. Draw this semicircle and label it as region R.

Step 2: Convert to polar coordinates

Now that we have a visualization of R, we will want to convert the given integral from Cartesian to polar coordinates. To do this, recall that \(x = r \cos(\theta)\), \(y = r \sin(\theta)\), and \(dA = r \ dr \ d\theta\). So the integral becomes: $$\iint_{R} 2 x y dA = \iint_{R} 2 (r \cos(\theta))(r \sin(\theta)) r \ dr \ d\theta$$ We also need to determine the limits of integration for r and theta. Since R is an upper semicircle with radius 3, the range for r is from 0 to 3 and for theta, it's from 0 to \(\pi\). So the integral becomes: $$\int_{0}^{\pi}\int_{0}^{3} 2 r^{3}( \sin(\theta)) (\cos(\theta)) dr d\theta$$

Step 3: Evaluate the integral with respect to r

Let's evaluate the inside integral first with respect to r: $$\int_{0}^{3} 2 r^{3} dr = \left[\frac{1}{2} r^{4}\right]_{0}^{3} = \frac{1}{2}(3^{4}) - \frac{1}{2}(0^{4}) = \frac{81}{2}$$ Now the integral becomes: $$\int_{0}^{\pi} \frac{81}{2} (\sin(\theta)) (\cos(\theta)) d\theta$$

Step 4: Evaluate the integral with respect to theta

To evaluate this integral, we will use the substitution \(u = \sin(\theta)\), so that \(du = \cos(\theta)d\theta\). The limits of integration for u will be 0 and 1, corresponding to the limits of theta from 0 to \(\pi\). $$\int_{0}^{1} \frac{81}{2} u du = \left[\frac{81}{4} u^{2}\right]_{0}^{1} = \frac{81}{4}$$

Step 5: Interpret the result

The value of the integral \(\iint_{R} 2xy dA\) over the region R is \(\frac{81}{4}\). This represents the sum of the function \(2xy\) over the specified region R confined by the upper semicircle with radius 3.

Key Concepts

Region of Integration

When approaching an integral problem, especially in a context requiring double integrals and a shift between coordinate systems, the **region of integration** is crucial. It's essentially the area or space over which the integration is carried out. In our exercise, the region of integration is defined by a set of inequalities: \(x^2 + y^2 \leq 9\) and \(y \geq 0\). This tells us that we're dealing with a semicircular region since the inequality \(x^2 + y^2 \leq 9\) forms a circle with radius 3 centered at the origin.

The condition \(y \geq 0\) means we are only interested in the upper half of this circle. Therefore, our region of integration is a semicircle on the Cartesian plane. Knowing the boundary and the conditions helps us sketch the area accurately, ensuring we capture the true limits of integration. Being able to visualize this helps enormously when setting up integral limits later on.

Convert Cartesian to Polar

A common task in calculus is to convert Cartesian coordinates to polar coordinates. This conversion can simplify the solving of integrals, especially when dealing with circular or radial symmetry. The expressions \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) are used to translate Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). Moreover, the element of area \(dA\) converts to \(r \ dr \ d\theta\) in polar coordinates.

The integral \(\iint_R 2xy \,dA\) becomes \(\iint_R 2(r \cos(\theta))(r \sin(\theta)) r \, dr \, d\theta\), combining what we know about \(x\) and \(y\) in polar terms. The next step is to determine limits for \(r\) and \(\theta\). For this semicircular region, \(r\) ranges from 0 to 3 (the radius), and \(\theta\) varies from 0 to \(\pi\) (covering the semicircle's upper half). By understanding these transformations, it becomes easier to perform the integration in a more manageable form.

Evaluate Double Integrals

Double integrals are daunting at first but become manageable with practice and structure, such as recognizing when to change coordinate systems or evaluate inner versus outer integrals. Once we have the double integral set up in polar coordinates, our task is to evaluate it by addressing each part in sequence.

For the integral \(\int_{0}^{\pi}\int_{0}^{3} 2r^3 \sin(\theta) \cos(\theta) \, dr \, d\theta\), we start with the inner integral. Calculating \(\int_{0}^{3} 2r^3 \, dr\), we find it equals \(\frac{81}{2}\). This value represents the integrated result over the radius from 0 to 3. We then multiply this by the remaining terms, \(\sin(\theta)\cos(\theta)\), and proceed to evaluate this simpler expression with respect to \(\theta\).

Using substitutions such as \(u = \sin(\theta)\), the integration becomes more straightforward. Ultimately, the result \(\frac{81}{4}\) gives a precise area or "sum" of the function \(2xy\) over the defined semicircular region R. This outcome reveals both the power and the necessity of double integrals in calculating areas and volumes where functions depend on more than one variable.