Question

Improper integrals Improper integrals arise in polar coordinates when the radial coordinate \(r\) becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$ \int_{\alpha}^{\beta} \int_{a}^{\infty} g(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} g(r, \theta) r d r d \theta $$ Use this technique to evaluate the following integrals. $$\iint_{R} \frac{d A}{\left(x^{2}+y^{2}\right)^{5 / 2}} ; R=\\{(r, \theta): 1 \leq r<\infty, 0 \leq \theta \leq 2 \pi\\}$$

Short answer

The value of the integral is $$-\frac{\pi}{4}$$.

Step-by-step solution

Convert to Polar Coordinates

We will convert the given integral into polar coordinates. Recall that \(x^2+y^2=r^2\). Also note that the differential area element \(dA\) in polar coordinates is given by \(r dr d\theta\). So, the given integral transforms to: $$\iint_R \frac{r dr d\theta}{(r^2)^{5/2}}$$ with the region being \((1\leq r<\infty\) and \(0\leq \theta \leq 2\pi)\).

Simplify the Integral

First simplify the integral by dividing the denominator with the numerator to get: $$\iint_R \frac{r dr d\theta}{(r^2)^{5/2}} = \iint_R r^{-9}dr d\theta$$

Calculate the First Integral

Now, we need to calculate the first integral with respect to r using the provided technique. We obtain: $$\lim_{b\rightarrow\infty} \int_1^b r^{-9} dr = \lim_{b\rightarrow\infty} \left[ \frac{-1}{8r^8}\right]_1^b$$

Evaluate the Limit

Evaluating the limit as \(b \rightarrow \infty\): $$\lim_{b\rightarrow\infty} \left[ \frac{-1}{8r^8}\right]_1^b = \frac{-1}{8} + 0 = \frac{-1}{8}$$

Calculate the Second Integral

Now, we have to compute the second integral with respect to θ: $$\int_0^{2\pi} \frac{-1}{8} d\theta$$

Evaluate the Second Integral

Evaluating the integral, we get: $$\int_0^{2\pi} \frac{-1}{8} d\theta = \frac{-1}{8} \left[\theta\right]_0^{2\pi} = -\frac{1}{8}(2\pi - 0) = -\frac{\pi}{4}$$ The final solution is: $$\iint_R \frac{dA}{(x^2+y^2)^{5/2}} = -\frac{\pi}{4}$$

Key concepts

Polar Coordinates

Polar coordinates offer a unique way to describe locations in a plane using the distance from a central point and an angle from a reference direction. Instead of using traditional Cartesian coordinates \(x, y\), polar coordinates use \(r, \theta\). \(r\) represents the radius, or how far a point is from the origin, and \(\theta\) is the angle in radians from the positive x-axis.
To convert between these two systems, use the formulas:

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

This method is particularly useful for problems involving circles or spirals, as these shapes are easily described in polar terms. In our exercise, transforming to polar coordinates simplifies the integration over a circular region.

Limit Evaluation

Limit evaluation comes into play when dealing with improper integrals, where the interval of integration goes to infinity. We use limits to ensure that integrals remain finite and meaningful.
For example, evaluating the integral:

• \(\lim_{b \rightarrow \infty} \int_1^b r^{-9} \, dr\)

First, find the antiderivative, and then take the limit as \(b\) approaches infinity.
This concept highlights how limits allow us to rigorously define and solve problems involving infinite intervals, controlling the behavior of functions that would otherwise diverge.

Multiple Integrals

Multiple integrals generalize the concept of integrating to handle functions of more than one variable, like in areas or volumes. In our exercise, we dealt with a double integral:

• \(\iint_R \, \frac{dA}{(x^2+y^2)^{5/2}}\)

This involves integrating first with respect to one variable (like \(r\)), and then another (like \(\theta\)). The order can often be switched depending on convenience or necessity. This process helps us find areas, volumes, and averages in higher dimensions, illustrating the power of calculus to tackle complex spatial problems.

Calculus Techniques

Various calculus techniques are employed to simplify and solve integrals. In our example, transforming to polar coordinates was essential, leading to a simplified problem.
Simplification techniques involve:

• Changing variables to match symmetry or shape of regions.
• Using identities or applying limits correctly.
• Breaking down complex integrals into manageable parts.

These strategies showcase the elegance of calculus, employing tools like substitution and partial fractions to handle even challenging integrals with ease, reinforcing the importance of flexibility and adaptability in mathematical problem-solving.