Question

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\begin{array}{l} \iint_{R} \frac{d A}{4+\sqrt{x^{2}+y^{2}}} ; R=\\{(r, \theta): 0 \leq r \leq 2 \\\ \pi / 2 \leq \theta \leq 3 \pi / 2\\} \end{array}$$

Short answer

Based on the step-by-step solution, evaluate the double integral: $$\iint_{R} \frac{dA}{4+\sqrt{x^{2}+y^{2}}},$$ where R is the region in the \(xy\)-plane enclosed by the polar curves \(r = 0\), \(r = 2\), \(\theta = \pi/2\), and \(\theta = 3\pi/2\). Answer: $$-\pi\ln\frac{3}{2}$$

Step-by-step solution

Convert the function to polar coordinates

We need to convert the given function from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). Recall the relationships between the two coordinate systems: $$x = r\cos\theta \quad\text{and}\quad y = r\sin\theta.$$ With these relations, we can rewrite the given function in polar coordinates, $$f(x, y) = \frac{1}{4 + \sqrt{x^2 + y^2}} = \frac{1}{4+\sqrt{(r\cos\theta)^2+(r\sin\theta)^2}} = \frac{1}{4+\sqrt{r^2(\cos^2\theta+\sin^2\theta)}} = \frac{1}{4+r}.$$ We also need to include the Jacobian for converting to polar coordinates, which is \(r\). This gives us the integral in polar coordinates: $$\iint_{R} \frac{1}{4+r} r\,drd\theta.$$

Set up the integral in polar coordinates

The region R, as given, is defined by: \(r\) ranges from \(0\leq r \leq 2\), \(\theta\) ranges from \(\pi / 2 \leq \theta \leq 3 \pi / 2\). Therefore, we can set up our double integral as follows: $$\int_{\pi/2}^{3\pi/2}\int_{0}^{2} \frac{r}{4+r}\,drd\theta.$$

Evaluation

We will first evaluate the integral with respect to \(r\) and then with respect to \(\theta\). Integrate with respect to \(r\): $$\int_{0}^{2} \frac{r}{4+r}\,dr = -\int_{4}^{6} \frac{1}{u}\,du$$ where we made the substitution \(u = 4+r\) and \(du = dr\). $$=-\left[\ln |u|\right]_{4}^{6} = -(\ln 6 - \ln 4) = \ln\frac{4}{6} = -\ln\frac{3}{2}.$$ Integrate with respect to \(\theta\): $$\int_{\pi/2}^{3\pi/2} -\ln\frac{3}{2}\,d\theta = -\ln\frac{3}{2}\left[\theta\right]_{\pi/2}^{3\pi/2} = -\ln\frac{3}{2}\left(\pi\right).$$ Hence, the double integral evaluates to: $$\iint_{R} \frac{dA}{4+\sqrt{x^{2}+y^{2}}} = -\pi\ln\frac{3}{2}.$$

Key concepts

Polar Coordinates

Polar coordinates provide a method to represent points in the plane using a radius and angle instead of the usual x and y coordinates. This system is particularly useful in situations involving circular or radial symmetry. In polar coordinates, a point is denoted as

• : the distance from the origin to the point, also known as the radial coordinate.
• \(\theta\): the angle measured from the positive x-axis to the line segment connecting the origin to the point, measured in radians.

To convert between Cartesian coordinates (x, y) and polar coordinates (r, \(\theta\)), we use:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

When performing integrations using polar coordinates, especially in cases involving circles or part of circles, it often simplifies the integration limits. This makes the process of evaluating certain integrals more straightforward once the function is converted appropriately.

Jacobian

The Jacobian is a critical factor when changing variables in multiple integrals. It is the determinant of the matrix of partial derivatives and serves as the scaling factor needed to convert the measurements when switching between different coordinate systems. In the case of converting from Cartesian coordinates to polar coordinates, the Jacobian \(r\) arises from the transformation, and it accounts for the change in area element. When integrating a function expressed in polar coordinates, the Jacobian tells us to incorporate an additional \(r\) term, transforming \(dA\) into \(r \, dr \, d\theta\). This accounts for the varying areas represented by the polar coordinate sections, maintaining the integrity of the integral.

Integration by Substitution

Integration by substitution is a technique used to make more complex integrals simpler. It involves changing variables to transform the integral into a more familiar or easily solvable form.In the given problem, after converting to polar coordinates, we used the substitution method on the inner integral when evaluating with respect to \(r\). The substitution \(u = 4 + r\) simplifies the \(\int \frac{r}{4+r} \, dr\) integral by turning it into the basic form \(\int \frac{1}{u} \, du\), a standard logarithmic integral. This highlights the versatility of substitution in handling non-trivial functional forms, transforming them into easier integrations.