Problem 30 - Thomas Calculus

Question

In Exercises \(21-30,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$ \int_{0}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} 6 x d y d x $$

Step-by-Step Solution

Answer

The reversed integral is \( \int_{-2}^{2} \int_{0}^{\sqrt{4 - y^2}} 6x \, dx \, dy \).

Step 1: Identify the Region of Integration

The given integral is \( \int_{0}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} 6x \, dy \, dx \). Here, \( x \) ranges from \( 0 \) to \( 2 \), and for each \( x \), \( y \) ranges from \( -\sqrt{4-x^{2}} \) to \( \sqrt{4-x^{2}} \). The equation \( y = \pm \sqrt{4 - x^2} \) represents the upper and lower halves of a semicircle with radius 2 centered on the origin. Therefore, the region is the right half of the circle \( x^2 + y^2 = 4 \) for \( x \geq 0 \).

Step 2: Sketch the Region

Draw the semicircle region on the coordinate plane. The semicircle centered at the origin with radius 2 is the circle equation \( x^2 + y^2 = 4 \), but we only consider \( x \ge 0 \). It forms the right semicircle extending from \( x = 0 \) to \( x = 2 \) and from \( y = -\sqrt{4-x^2} \) to \( y = \sqrt{4-x^2} \).

Step 3: Determine the Bounds for Flipping the Integral

Since \( y \) is expressed as \( \pm \sqrt{4-x^2} \), it ranges from \( -2 \) to \( 2 \) for the entire semicircle. For every fixed \( y \), \( x \) ranges from \( 0 \) to \( \sqrt{4-y^2} \). The variable \( x \) goes from the y-axis out to the semicircle boundary.

Step 4: Write the Reversed Integral

The reversed order of integration would integrate over \( y \) first and then \( x \). Therefore, the equivalent double integral with reversed bounds is: \( \int_{-2}^{2} \int_{0}^{\sqrt{4 - y^2}} 6x \, dx \, dy \).

Key Concepts

Order of Integration

In double integrals, the order of integration refers to the sequence in which we integrate with respect to the variables. Whenever you have a double integral, you will notice a setup like \( \int_a^b \int_c^d f(x,y) \, dy \, dx \). Here, the integration takes place over two variables—in this case, \(y\) and \(x\). The integration with respect to \(y\) (\(dy\)) occurs first, followed by \(x\) (\(dx\)). This sequence is important because it determines how the region of integration in the \(xy\)-plane is traced and calculated.
Switching the order of integration often simplifies the integration process. For example, if integrating with respect to \(y\) first becomes too cumbersome, we switch and start with \(x\). This requires careful adjustment of integration limits to correctly represent the same region in the reversed setup. In essence, the order of integration involves deciding the path of integration through a given region.

Region of Integration

In order to set up a double integral correctly, understanding the region of integration is essential. In the given exercise, the region is defined by the integral limits: \(\int_0^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 6x \, dy \, dx\).

• The limits for \(x\), ranging from 0 to 2, define the outer boundary along the horizontal axis.
• The limits for \(y\), ranging from \(-\sqrt{4-x^2}\) to \(\sqrt{4-x^2}\), create a vertical boundary related to a semicircle.

The specific area you're integrating over is the right half of a circle, where each value of \(x\) corresponds to values of \(y\) that satisfy the circle equation \(x^2 + y^2 = 4\), but for \(x \geq 0\). Visualizing this specific region ensures you're correctly computing the integral—miss this, and your results won't match reality.

Semicircle in Integration

Semicircles frequently emerge in mathematical problems involving integration, especially in cases with circular symmetry like those encountered in physics or engineering. In the given problem, we identify the region of integration as half a circle (a semicircle). This specific semicircle arises from the equation \(y = \pm \sqrt{4-x^2}\), representing a circle of radius 2.
As the problem defines, this semicircle covers all points where \(x^2 + y^2 = 4\), with \(x \geq 0\). Visualizing this can mean sketching the right side of the circle, set against the \(y\)-axis, extending from \(y = -2\) to \(y = 2\).
When setting up a double integral over a semicircular region, it's crucial to understand both the radial symmetry and boundary constraints. Here, by reversing the integration order, we switch to integrating vertically first (over \(y\)) and horizontally second (over \(x\)), ideally capturing the same geometric region but potentially simplifying calculation.