Question

Miscellaneous integrals Sketch the region of integration and evaluate the following integrals, using the method of your choice. $$\int_{-4}^{4} \int_{0}^{\sqrt{16-y^{2}}}\left(16-x^{2}-y^{2}\right) d x d y$$

Short answer

Based on the step by step solution provided, determine the value of the double integral. Answer: \(-\frac{29696}{3}\)

Step-by-step solution

Sketch the region of integration

To sketch the region of integration, we need to find the relationship between the bounds of x and y. The inner integral has the bounds \(0 \leq x \leq \sqrt{16-y^{2}}\). The outer integral has the bounds \(-4 \leq y \leq 4\). The inner integral gives the equation of a semicircle with radius 4: $$x^2 + y^2 = 16$$ The region of integration is the part of this semicircle that lies in the first and the fourth quadrants (\(x\ge0\)).

Integrating with respect to x

We will first integrate the function with respect to x: $$\int_{0}^{\sqrt{16-y^{2}}}\left(16-x^{2}-y^{2}\right) dx$$ To find the antiderivative with respect to x, we have: $$\left[16x - \frac{1}{3}x^3 - y^2x\right]_{0}^{\sqrt{16-y^{2}}}$$ Using the fundamental theorem of calculus, we get: $$\left[16\sqrt{16-y^2} - \frac{1}{3}\left(16-y^2\right)^{\frac{3}{2}} - y^2\sqrt{16-y^2}\right] - \left[0\right]$$ This simplifies to: $$16\sqrt{16-y^2} - \frac{1}{3}\left(16-y^2\right)^{\frac{3}{2}} - y^2\sqrt{16-y^2}$$

Integrating with respect to y

Now, we will integrate the result with respect to y over the limits -4 to 4: $$\int_{-4}^{4} \left(16\sqrt{16-y^2} - \frac{1}{3}\left(16-y^2\right)^{\frac{3}{2}} - y^2\sqrt{16-y^2}\right) dy$$ To evaluate this integral, we can use a substitution. Let: $$u = 16-y^2 \Rightarrow -2ydy = du$$ The limits of integration now become: $$u(-4) = 0, u(4) = 16$$ Now, the integral becomes: $$\int_{16}^{0} \left(16\sqrt{u} - \frac{1}{3}u\sqrt{u} -\frac{-u\sqrt{u}}{-4}\right) \frac{-1}{2}\frac{du}{\sqrt{u}}$$ Simplify the integral: $$\frac{1}{2}\int_{16}^{0} \left(32\sqrt{u} - \frac{2}{3}u - u\right) du$$ To find the antiderivative with respect to u, we have: $$\frac{1}{2}\left[\frac{32}{3}u^{\frac{3}{2}} - \frac{1}{3}u^2 - \frac{1}{2}u^2\right]_{16}^{0}$$ Using the fundamental theorem of calculus, we get: $$\frac{1}{2}\left[\frac{32}{3}(0)^{\frac{3}{2}} - \frac{1}{3}(0)^2 - \frac{1}{2}(0)^2\right] - \frac{1}{2}\left[\frac{32}{3}(16)^{\frac{3}{2}} - \frac{1}{3}(16)^2 - \frac{1}{2}(16)^2\right]$$ Simplifying the expression, we obtain the final value of the integral: $$-\frac{1}{2}\left[\frac{32768}{3} - \frac{256}{3} - 128\right] = \boxed{-\frac{29696}{3}}$$

Key concepts

Region of Integration

The first step in solving a double integral problem is to understand the region over which we are integrating. In this exercise, we're given the double integral \[\int_{-4}^{4} \int_{0}^{\sqrt{16-y^{2}}}(16-x^{2}-y^{2})\, dx \, dy\].
The inner integral has bounds defined by \(0 \leq x \leq \sqrt{16-y^{2}}\), which describe a semicircle since when you rearrange the bounds, you have the equation \(x^2 + y^2 = 16\). This is the equation of a circle with a radius of 4. Therefore, the region of integration is a semicircle centered at the origin with a radius of 4, restricted to the right of the y-axis, or in other words, where \(x \geq 0\).
The outer integral has bounds \(-4 \leq y \leq 4\), meaning we're considering all y-values from -4 to 4.
Together, these constraints describe the region of integration. Understanding this region allows us to change variables systematically and correctly, as we process through the solving method.

Polar Coordinates

Using polar coordinates is advantageous when dealing with circular symmetry or radial bounds like the ones found in semicircular regions.
In polar coordinates, points in the plane are represented with angles and distances from the origin—specifically, \((r, \theta)\)—where \(r\) is the radius and \(\theta\) is the angle.
Converting to polar coordinates in this problem, the equation \(x^2 + y^2 = 16\) becomes \(r^2 = 16\), or \(r = 4\), which confirms the circular nature of our region.
When converting an integral from Cartesian \((x, y)\) coordinates to polar coordinates, we use the transformations:

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

The differential area element, \(dx\, dy\), is replaced by \(r \, dr \, d\theta\). This helps make the integration process much simpler when evaluating areas with circular symmetry.
In polar coordinates, the limits for \(r\) would be from 0 to 4, and \(\theta\) would extend from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) to cover our specific semicircle in the correct quadrants.

Substitution Method

The substitution method is a powerful technique used to simplify the process of integration by transforming the integral into a new domain with different variables.
In this exercise, after the initial integration with respect to \(x\), the remaining integral is simplified using substitution.
The substitution \(u = 16 - y^2\) is introduced to transform the inner integral, allowing us to handle the more complex part of the expression analytically.
The differential transformation follows as \(du = -2y \, dy\). This implies that the integration variable \(y\) and its half \(dy\), when combined, transform into separate variables \(u\) and \(du\), making the integration feasible.
Changing the limits of integration is crucial with substitution, so the original \(y\)-limits \(-4\) to \(4\) transform to \(u(4) = 0\) to \(u(-4) = 16\). This simplifies and corrects the integral effectively, optimizing the integration process.
By substituting, the complex square roots and expressions inside the integral become easier, well-defined polynomials, which are then direct to integrate.