Question

Miscellaneous integrals Sketch the region of integration and evaluate the following integrals, using the method of your choice. $$\begin{array}{l} \iiint_{R} \frac{1}{4+\sqrt{x^{2}+y^{2}}} d A ; R=\\{(r, \theta): 0 \leq r \leq 2 \\ \pi / 2 \leq \theta \leq 3 \pi / 2\\} \end{array}$$

Short answer

Question: Calculate the value of the triple integral over the given region R, defined by 0 ≤ r ≤ 2, and π/2 ≤ θ ≤ 3π/2, for the function f(x,y) = (1/(4 + √(x² + y²))). Express your answer as a simplified expression. Answer: The value of the triple integral over region R for the given function is ln(3/2)π.

Step-by-step solution

Sketch region of integration

First, let's sketch the given region of integration R. In polar coordinates, it's defined by: - 0 ≤ r ≤ 2 - π/2 ≤ θ ≤ 3π/2 This is a portion of a circle with a radius equal to 2, spanning from the positive y-axis to the negative y-axis, in the second and third quadrants.

Convert the integrand to polar coordinates

Next, we need to convert the given function into polar coordinates. We know that: - x = r cos(θ) - y = r sin(θ) So, the integrand can be expressed as: $$\frac{1}{4+\sqrt{(r\cos{\theta})^2+(r\sin{\theta})^2}}$$ Which simplifies to: $$\frac{1}{4+\sqrt{r^2}}$$

Set up the triple integral

Now, we can set up the triple integral. Don't forget the Jacobian factor r when converting to polar coordinates: $$\iiint_{R} \frac{1}{4+\sqrt{r^2}}r\,dr\,d\theta$$ Let's find the antiderivative with respect to r and evaluate it over the given limits.

Evaluate the inner integral with respect to r

First, evaluate the inner integral: $$\int_{0}^{2} \frac{r}{4+\sqrt{r^2}}\,dr$$ To solve this integral, use the substitution \(u = 4 + \sqrt{r^2}\) with \(du/dr = (r/\sqrt{r^2})\). So \(r dr = du\). After applying the substitution, we get: $$\int \frac{1}{u}\,du$$ Now, simply evaluate this integral: $$\ln(u) + C = \ln(4 + \sqrt{r^2}) + C$$ Now, applying the limits of integration for r, we get: $$[\ln(4 + \sqrt{r^2})]_0^{2} = (\ln(6) - \ln(4)) = \ln(\frac{3}{2})$$

Evaluate the outer integral with respect to θ

Now, we will evaluate the outer integral: $$\int_{\pi/2}^{3\pi/2} \ln(\frac{3}{2})\,d\theta$$ This is a simple integral since \(\ln(\frac{3}{2})\) is just a constant: $$\ln(\frac{3}{2})[\theta]_{\pi/2}^{3\pi/2} = (\ln(\frac{3}{2}))(3\pi/2 - \pi/2) = \ln(\frac{3}{2})\pi$$

Final answer

So, the result of the triple integral is: $$\iiint_{R} \frac{1}{4+\sqrt{x^{2}+y^{2}}} dA = \ln(\frac{3}{2})\pi$$

Key concepts

Region of Integration

In polar coordinates, a region of integration represents the specific area over which an integral is evaluated. It dictates the limits within which the variables, typically the radius \(r\) and the angle \(\theta\), must remain. For our exercise, the region of integration is defined by two inequalities: \(0 \leq r \leq 2\) and \(\pi/2 \leq \theta \leq 3\pi/2\). These parameters describe a sector of a circle with a radius of 2.
The angle \(\theta\) extends from \(\pi/2\) to \(3\pi/2\), which spans the second and third quadrants of the Cartesian plane. This spans clockwise from the positive y-axis, countering into the negative y direction. Hence, the region forms a semicircle, providing a clear boundary for where the integration should occur.
Understanding this region is crucial for setting up the subsequent limits of integration in polar coordinate integrals.

Jacobian Factor

When transforming an integral from Cartesian to polar coordinates, the Jacobian factor is essential. It accounts for the change in area element caused by the transformation. In polar coordinates, the differential area, originally expressed as \(dA = dx \, dy\), becomes \(dA = r \, dr \, d\theta\).
The factor \(r\) is what we call the Jacobian factor. It compensates for the distortion that occurs when transforming a small area from a rectangular shape (Cartesian) to a wedge-like shape (polar).

• The Jacobian factor is crucial to ensure that the transformed integral accurately represents the same magnitude as it would in Cartesian coordinates.
• It's a simple multiplication by \(r\), but this is not a minor adjustment. Skipping or incorrectly applying the Jacobian would result in an incorrect integral value since it considers how areas scale in the new coordinate system.

This small factor encapsulates how transformation alters elements of integration.

Polar Coordinates Transformation

Polar coordinates transformation involves changing from a Cartesian coordinate system \((x, y)\) to a polar system \((r, \theta)\). This transformation is particularly useful for circular or radial symmetry problems, making the integration process simpler and more intuitive.
The transformation is guided by the relations:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(r = \sqrt{x^2 + y^2}\)
• \(\theta = \tan^{-1}(y/x)\)

In our problem, the original integrand \(\frac{1}{4+\sqrt{x^2+y^2}}\) is rewritten as \(\frac{1}{4+\sqrt{r^2}}\) by this transformation.
The transformation drastically simplifies problems involving regions bounded by circles or other symmetrical shapes. It's not just a convenience but often a necessity when integrating over circular regions.