Question

Miscellaneous integrals Sketch the region of integration and evaluate the following integrals, using the method of your choice. $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x$$

Short answer

Question: Sketch the region of integration for the given double integral and evaluate the integral using any method: $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} dy dx$$ Answer: The region of integration is a semi-circle with a radius of 3 centered at the origin on the right side of the y-axis. The value of the double integral is \(9\pi\).

Step-by-step solution

Sketch region of integration

Consider \(0 \leq y \leq \sqrt{9-x^2}\). This can be rewritten as \(x^2+y^2 \leq 9\). This inequality represents the area contained within a circle of radius 3 centered at the origin. Furthermore, \(0 \le x \le 3\), so we are interested only in the right half of the circle. Thus, the region of integration is a semi-circle with radius 3 centered at the origin on the right side of the y-axis.

Convert to polar coordinates

To integrate the given function more easily, we can convert it to polar coordinates. Recall that \(x = r\cos{\theta}\), \(y = r\sin{\theta}\), and \(dA = r\,dr\,d\theta\). With these substitutions, our integral becomes: $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} dy dx = \int_{0}^{\pi}\int_{0}^{3}r^2 dr d\theta$$ The bounds for \(r\) go from \(0\) to \(3\), and the bounds for \(\theta\) go from \(0\) to \(\pi\) since we are dealing with a semi-circle in the positive \(x\) direction.

Evaluate the integral

Now we will evaluate the integral: $$\int_{0}^{\pi}\int_{0}^{3}r^2 dr d\theta = \int_{0}^{\pi} \left[ \frac{r^3}{3}\right]_{0}^{3} d\theta = \int_{0}^{\pi} \frac{27}{3} d\theta = 9 \int_{0}^{\pi} d\theta$$ Finally, integrating with respect to \(\theta\), we get: $$9 \int_{0}^{\pi} d\theta = 9 [\theta]_{0}^{\pi} = 9(\pi - 0) = 9\pi$$ Thus, the value of the double integral is \(9\pi\).

Key concepts

Polar Coordinates

Polar coordinates provide an alternative way to describe points in a plane using a pair of values: the radial distance from the origin and the angle from the positive x-axis. This system is particularly useful in cases where the region of interest is circular or involves radial symmetry.
In polar coordinates, each point in the plane is identified by

• the radius: \(r\), which is the distance from the origin, and
• the angle: \(\theta\), which is measured from the positive x-axis counterclockwise.

The conversion between Cartesian coordinates \((x, y)\) and polar coordinates \((r, \theta)\) involves the equations:

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

Switching to polar coordinates is advantageous in integrating over circular regions, as exemplified in our original exercise. The transformation simplifies the expression \(\sqrt{x^2 + y^2}\) to just \(r\), and the differential area element \(dA\) becomes \(r\,dr\,d\theta\). This makes integration more straightforward over circular areas.

Region of Integration

The region of integration refers to the specific area over which a double integral is to be evaluated. It is important to accurately define this region to properly evaluate the integral.
In the original exercise, we are given the bounds \(0 \leq y \leq \sqrt{9-x^2}\) and \(0 \leq x \leq 3\). To understand this region, we note that the equation \(y = \sqrt{9-x^2}\) forms a semi-circle on the graph because \(x^2 + y^2 = 9\) is a circle equation with radius 3.
Since the exercise specifies \(0 \leq x \leq 3\), we focus on the right-half of this circle, where \(x\) is positive. As a result, our region of integration is a semi-circle centered on the origin, lying in the upper right quadrant of the Cartesian plane.

Integration Bounds

Integration bounds are limits that define the range over which integration occurs. When dealing with double integrals, it is crucial to set these limits correctly based on the region of integration.
To transform to polar coordinates, the bounds for \(r\) and \(\theta\) must reflect the original region. In our exercise, the region is a semi-circle of radius 3, centered at the origin, in the right half of the plane. This influences how we choose the bounds.

• The bounds for \(r\) are \(0\) to \(3\) because \(r\) measures the distance from the origin to any point on the circumference of the circle.
• The bounds for \(\theta\) are \(0\) to \(\pi\) because this angle range covers the upper right semi-circle. It starts from the positive x-axis and winds counterclockwise to completely envelop the right half of the circle.

Correctly establishing these integration bounds ensures that the integration covers the entire intended region, leading to an accurate evaluation of the double integral.