Question

Miscellaneous integrals Sketch the region of integration and evaluate the following integrals, using the method of your choice. $$\int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\left(x^{2}+y^{2}\right)^{3 / 2} d y d x$$

Short answer

Answer: The value of the double integral is (2π/5).

Step-by-step solution

Sketch the region of integration

To visualize the region of integration, plot the boundary curves on the Cartesian plane. The limits of integration for \(x\) are from \(-1\) to \(1\). The limits of integration for \(y\) involve the functions \(y=\sqrt{1-x^2}\) and \(y=-\sqrt{1-x^2}\), which represent the upper and the lower semicircle of a unit circle centered at the origin. The region of integration lies within this unit circle.

Choose the method of integration

The function \((x^2 + y^2)^{3/2}\) is symmetric with respect to the \(x\)-axis and the region of integration has circular symmetry. Consequently, it is easier to evaluate the double integral using polar coordinates. To this end, we need to convert the integral from Cartesian to polar coordinates by performing the following substitutions: - \(x = r\cos(\theta)\) - \(y = r\sin(\theta)\) - \(dA = r dr d\theta\) The integral becomes: $$\int_{0}^{2\pi}{\int_{0}^{1} (r^2\cos^2(\theta) + r^2\sin^2(\theta))^{3/2} \cdot r dr d\theta}.$$

Simplify the integrand

Now, simplify the integrand as follows: \((r^2\cos^2(\theta) + r^2\sin^2(\theta))^{3/2} \cdot r = r^4\) The integral becomes: $$\int_{0}^{2\pi}{\int_{0}^{1} r^4 dr d\theta}.$$

Evaluate the inner integral

Evaluate the inner integral with respect to r: $$\int_{0}^{1} r^4 dr = \frac{1}{5}r^5\Big|_0^1 = \frac{1}{5}.$$ Now the integral is: $$\int_{0}^{2\pi}{\frac{1}{5} d\theta}.$$

Evaluate the outer integral

Evaluate the outer integral with respect to \(\theta\): $$\int_{0}^{2\pi}{\frac{1}{5} d\theta} = \frac{1}{5}\theta\Big|_0^{2\pi} = \frac{2\pi}{5}.$$ Thus, the value of the double integral is \(\frac{2\pi}{5}\).

Key concepts

Polar Coordinates

When working with integrals over circular regions, polar coordinates can simplify the process. Unlike Cartesian coordinates, which use \(x\) and \(y\), polar coordinates are expressed as \(r\) (the distance from the origin) and \(\theta\) (the angle from the positive \(x\)-axis).
This system is particularly useful for dealing with circles and circular symmetries, as

• Circular boundaries become constant-radius lines, making them easier to handle.
• The area element \(dA\) in polar coordinates is \(r \, dr \, d\theta\), reflecting the radial increments.

In terms of conversion, we use the formulas:

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

These conversions are key to transitioning from Cartesian areas to their simpler polar counterparts, especially when integrating over circular regions.

Conversion of Integrals

Switching an integral from Cartesian to polar coordinates can make calculations simpler, particularly when dealing with circular symmetries. The original integral in Cartesian form can often be complex and challenging to solve directly. By converting, you leverage the symmetry of circular regions.
The conversion involves substituting:

• The expressions \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).
• The area element \(dA = dxdy\) becomes \(r \, dr \, d\theta\).

After conversion, our integrand \((x^2 + y^2)^{3/2}\) simplifies significantly, often reducing complex expressions to simpler ones involving \(r\) and \(\theta\). This substitution makes it easier to assess the true nature of the function's growth and its impact across the circular domain.

Region of Integration Visualization

Visualizing the region of integration is a crucial step to understanding the scope of an integral. For the given exercise, plotting the boundaries using the equations \(y = \sqrt{1-x^2}\) and \(y = -\sqrt{1-x^2}\) reveals a region shaped like a complete circle.
This circle is centered at the origin with a radius of 1 (hence the use of a unit circle). By visualizing:

• We confirm that the integrals' limits correspond to a well-defined, symmetric region.
• It becomes apparent why switching to polar coordinates, where \(r\) varies from 0 to 1 and \(\theta\) varies from 0 to \(2\pi\), is beneficial.

Understanding how these shapes translate between Cartesian and polar systems helps in realizing how certain coordinate systems can greatly simplify the process of integration.

Symmetry in Integration

Symmetry plays a pivotal role in simplifying integration processes. In many integrals, recognizing symmetry can significantly reduce computation time by simplifying complex areas in higher dimensions.
In our integral, the symmetry of the circle around the origin makes it particularly easy to handle. Here are some key points:

• The function \((x^2 + y^2)^{3/2}\) demonstrates radial symmetry due to its expression, where \(x^2 + y^2 = r^2.\)
• The region itself (a full circle) supports this symmetry, allowing us to more efficiently set up and evaluate the integral.

Utilizing symmetry means you may only compute one portion of the area and infer the whole by these symmetrical properties, as we saw with the initial choice to use polar coordinates regarding this circle. Symmetrical attributes are not only elegant but also highly practical in calculations.