Problem 55 - Calculus Early Transcendentals

Question

Find the area of the regions bounded by the following curves. \text { The lemniscate } r^{2}=6 \sin 2 \theta

Step-by-Step Solution

Answer

Answer: The area of the region bounded by the curve is \(\frac{3}{2}\) square units.

Step 1: Identify limits of integration

Observe that the lemniscate is symmetric with respect to the origin. To find the limits of integration, set \(r^2\) equal to zero and solve for \(\theta\). \[r^2 = 6\sin 2\theta \Longrightarrow 0 = 6\sin 2\theta\] Divide both sides by 6: \[\sin 2\theta = 0\] Solving for \(\theta\), we get \(\theta = 0, \frac{\pi}{4}, \frac{\pi}{2}\). Therefore, we can integrate from \(0\) to \(\frac{\pi}{4}\) to find the area of one lobe.

Step 2: Set up and compute the integral for one lobe

To find the area of one region, we will use the following formula for the area under the curve in polar coordinates: \[A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta\] In this case, \(\alpha = 0\) and \(\beta = \frac{\pi}{4}\). Plug in the equation for \(r^2\): \[A = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} 6\sin 2\theta d\theta\] Now, integrate: \[A = \frac{3}{2} \left[-\frac{1}{2} \cos 2\theta \right]_{0}^{\frac{\pi}{4}}\]

Step 3: Evaluate the integral for one lobe

substitution of the limits of integration: \[A = \frac{3}{2} \left[-\frac{1}{2} \cos \frac{\pi}{2} - (-\frac{1}{2}\cos 0)\right]\] \[A = \frac{3}{2} \left[0 +\frac{1}{2}\right] = \frac{3}{4}\]

Step 4: Calculate the area of the whole lemniscate

Since we found the area for one lobe, we can now calculate the total area by multiplying the area of one lobe by 2 (due to symmetry): \[A_\text{total} = 2 \cdot \frac{3}{4} = \frac{3}{2}\] The area of the regions bounded by the lemniscate is \(\frac{3}{2} \text{ square units}\).

Key Concepts

Understanding the Lemniscate

The lemniscate is a fascinating geometric figure that resembles a figure-eight or an infinity symbol. In polar coordinates, this particular shape is characterized by the equation \( r^2 = 6 \sin 2 \theta \).
This equation helps define the distance \( r \) from the origin, depending on the angle \( \theta \).

• A lemniscate is symmetric around the origin, meaning that both halves are mirror images.
• The symmetry can be leveraged to simplify calculations, such as finding the area enclosed by the curve.
• The sine function within the equation indicates that the shape will repeat itself with every full rotation \( (0 - 2\pi) \).

Understanding the properties of a lemniscate in polar form is crucial for solving problems related to its area and integration limits.

Calculating the Area of a Region

Finding the area of a region enclosed by a curve in polar coordinates involves integrating over a defined angle range.
The general formula to calculate the area \( A \) for a curve \( r(\theta) \) is:\[A = \frac{1}{2} \int_{\alpha}^{\beta} r^2\, d\theta\]

• The \( \frac{1}{2} \) factor comes from the way area is accumulated in polar coordinates, unlike the Cartesian system.
• For a lemniscate with the equation \( r^2 = 6 \sin 2 \theta \), our task is to integrate \( r^2 \) over one lobe.
• By computing the integral from \( \theta = 0 \) to \( \theta = \frac{\pi}{4} \), we get the area of one loop. This is due to the shape's symmetric nature.

Once you have the area of one lobe, the total area can be found by doubling the result because the lemniscate has two symmetric loops.

Defining Integration Limits

When working with polar coordinates, setting the correct integration limits is vital to accurately determining the area.
This involves identifying where the curve crosses itself or closes, which is often where the radius \( r = 0 \) or where its periodic nature repeats.
For the lemniscate \( r^2 = 6 \sin 2 \theta \), setting \( r^2 = 0 \) helps find the angles \( \theta \) that form the integration limits:

• Solving \( \sin 2\theta = 0 \) gives us the values for \( \theta \) where our curve begins and ends.
• This results in \( \theta = 0 \), \( \theta = \frac{\pi}{4} \), and further angles that denote comprehensively covering the entire range of the shape.

Because of their periodicity, trigonometric functions like sine can limit integrations to one period or lobe, making calculations simpler and more efficient.
Remember to reflect on symmetry properties to decide whether to calculate for one section and multiply, or integrate the full range at once for complete accuracy.