Question

Sketch the region of integration and evaluate the following integrals as they are written. $$\int_{0}^{1} \int_{-2 \sqrt{1-y^{2}}}^{2 \sqrt{1-y^{2}}} 2 x d x d y$$

Short answer

Question: Determine the value of the given double integral: $$\int_{0}^{1} \int_{-2 \sqrt{1-y^{2}}}^{2 \sqrt{1-y^{2}}} 2x dxdy$$ Answer: The value of the given double integral is \(0\).

Step-by-step solution

Sketch the Region of Integration

To start sketching the region of integration, let's analyze the outer integral with respect to \(y\): $$ 0 \le y \le 1 $$ This simply means that the region is limited between \(y=0\) and \(y=1\). Now let's analyze the inner integral with respect to \(x\): $$ -2\sqrt{1-y^2} \le x \le 2\sqrt{1-y^2} $$ These bounds represent two semi-circle with \(y\) ranging from \(0\) to \(1\). The left semi-circle is centered at \(x=0\) with radius \(2\). So, our region of integration is the area enclosed within these semi-circles and \(y=1\).

Evaluate the Inner Integral

Now, let's evaluate the inner integral with respect to \(x\): $$ \int_{-2\sqrt{1-y^2}}^{2\sqrt{1-y^2}} 2x dx $$ This is simply the integral of \(2x\) with the given limits. Integrating, we get: $$ \left. x^2\right|_{-2\sqrt{1-y^2}}^{2\sqrt{1-y^2}} $$ Now, evaluating the expression at the limits, we get: $$ (2\sqrt{1-y^2})^2 - (-2\sqrt{1-y^2})^2 = 4(1-y^2) - 4(1-y^2) = 0 $$

Evaluate the Outer Integral

Since the result of the inner integral evaluation is \(0\), the entire integral becomes: $$ \int_{0}^{1} 0 dy $$ Evaluating this integral is simple, as the integrand is the constant \(0\). The result is also \(0\): $$ \left. 0y\right|_0^1 = 0 $$ So, the value of the given double integral is \(0\).

Key concepts

Region of Integration

Understanding the **region of integration** is crucial for solving double integrals. This region defines the area over which the integral is calculated. It combines the boundaries of both inner and outer integrals.
In the given exercise, the outer integral has boundaries for the variable \(y\) ranging from 0 to 1. This establishes vertical limits for the region. Meanwhile, the inner integral's limits, \(-2\sqrt{1-y^2} \leq x \leq 2\sqrt{1-y^2}\), describe a horizontal range that resembles two semicircular arcs. These are centered along the \(x\)-axis and stretch outwards with a radius of 2.

• The parameter \(x\) changes with \(y\), forming curves instead of straight lines typical for constant boundaries.
• The boundary conditions jointly define a quarter of the disk, limited to the first quadrant, with \(y\) moving from 0 to 1.

Visualizing this region helps to better understand what area the double integral covers and why.

Polar Coordinates

**Polar coordinates** offer a useful alternative to Cartesian coordinates, especially when dealing with circular or radial symmetries. They express points in terms of radial distance \(r\) and angle \(\theta\). Often, switching to polar coordinates simplifies the limits of integration and the integrand itself.
In our exercise, while not directly necessary, using polar coordinates could help visualize the semicircles described by \(x = \pm 2\sqrt{1-y^2}\). These semicircles correspond to circles in polar coordinates, offering symmetry that's sometimes easier to handle:

• Transform \(x\) and \(y\) using \(x = r\cos\theta\) and \(y = r\sin\theta\).
• Re-express \(x^2 + y^2 = r^2\) to correspond with boundaries defined by the radius.
• Integrate over \(r\) and \(\theta\) instead of \(x\) and \(y\), sometimes simplifying calculations.

Polar transformations might not simplify every problem but can be particularly powerful with radial symmetries.

Iterated Integrals

**Iterated integrals** break the double integral down into a series of simpler, single-variable integrals. This technique involves evaluating each integral step by step, often starting from the innermost one.
In the exercise, calculating the double integral involves iterating over variables one at a time:

• First, compute the inner integral \(\int_{-2\sqrt{1-y^2}}^{2\sqrt{1-y^2}} 2x \, dx\). This defines how \(x\) varies relative to \(y\).
• Next, evaluate the outer integral \(\int_{0}^{1} \, 0 \, dy\), using the result of the inner integral.

This approach effectively manages complex expressions by reducing them to parts that are easier to assess individually.
Understanding iterated integrals is essential for solving double integrals efficiently. It highlights the importance of integrating in order and correctly applying limits for each variable.