Problem 69 - Thomas Calculus

Question

Find the areas of the regions enclosed by the lines and curves in Exercises \(63-70 .\) $$ x=3 \sin y \sqrt{\cos y} \quad \text { and } \quad x=0, \quad 0 \leq y \leq \pi / 2 $$

Step-by-Step Solution

Answer

The area enclosed is 2 square units.

Step 1: Understand the problem

We have a curve given by the equation \(x = 3\sin y \sqrt{\cos y}\) and a vertical line \(x = 0\). We need to find the area enclosed by these lines and curves from \(y = 0\) to \(y = \pi/2\). The region lies on the y-axis since \(x = 0\) is one of the boundaries.

Step 2: Set up the integral

The area enclosed can be found by integrating the function \(x(y) = 3\sin y \sqrt{\cos y}\) with respect to \(y\). Since \(x\) ranges from 0 to the curve, we can write the area integral as: \[A = \int_{0}^{\pi/2} x(y) \; dy = \int_{0}^{\pi/2} 3\sin y \sqrt{\cos y} \; dy\]

Step 3: Solve the integral using substitution

Letting \(u = \cos y\) changes the differential \(dy = -\frac{1}{\sin y} du\). Also, note that when \(y = 0\), \(u = \cos(0) = 1\), and when \(y = \pi/2\), \(u = \cos(\pi/2) = 0\). Rewrite the integral: \[A = \int_{1}^{0} 3\sqrt{u} \; (-\frac{1}{\sin y}) \, du = 3 \int_{0}^{1} \sqrt{u} \; du\]

Step 4: Evaluate the integral

The integral \(\int \sqrt{u} \; du\) can be rewritten using the power rule as \(\int u^{1/2} \; du = \frac{2}{3}u^{3/2}\). Now evaluate from 0 to 1:\[A = 3 \times \left[ \frac{2}{3}u^{3/2} \right]_{0}^{1} = 3 \times \left( \frac{2}{3}(1) - \frac{2}{3}(0) \right) = 3 \times \frac{2}{3} = 2\]

Step 5: Conclusions

The area of the region enclosed by the curve \(x = 3\sin y \sqrt{\cos y}\) and \(x = 0\) from \(y = 0\) to \(y = \pi/2\) is \(2\) square units.

Key Concepts

Definite Integral

A definite integral represents the signed area under a curve between two specific limits. In this exercise, we are interested in finding the area enclosed between the curve and the y-axis. The definite integral provides a way to calculate this area precisely.

The function given is expressed in terms of a variable, here it's in the form of a trigonometric expression, involving both sine and cosine. To solve it, we need to understand how the function behaves between the specified limits. For our curve, the limits are from \(y = 0\) to \(y = \pi/2\).

By setting up the definite integral, we can compute the total area where the curve lies above the y-axis within these limits:

• The curve equation is \(x(y) = 3\sin y \sqrt{\cos y}\), making the x-coordinate a function of y.
• The integral setup reads \(\int_{0}^{\pi/2} 3\sin y \sqrt{\cos y} \; dy\), which directly gives us the enclosed area.

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify the integration of expressions involving trigonometric functions. In this problem, the curve function involves both sine and cosine, making it a likely candidate for this method.

A trigonometric substitution aims to change the variable in the integral to make it easier to solve. Here, we use the substitution \(u = \cos y\), which is chosen because the differentiation of \(\cos y\) leads to the appearance of \(\sin y\) in the differential, which matches part of our original integral:

• When \(u = \cos y\), the differential becomes \(dy = -\frac{1}{\sin y} du\).
• By changing limits: when \(y = 0\), \(u = 1\), and when \(y = \pi/2\), \(u = 0\).
• The substitution allows the integral to transform into a simpler form: \(3 \int_{1}^{0} \sqrt{u} \; (-\frac{1}{\sin y}) \, du\).

This results in a more straightforward integral that can be easily evaluated.

Integration Techniques

Integration techniques are critical in evaluating the integrals once they have been properly set up through methods like trigonometric substitution.

After the substitution in this problem, the integral becomes \(3 \int_{0}^{1} \sqrt{u} \, du\). This falls into a category of integrals where the power rule is applicable. The power rule for integration states that the integral of \(u^n\) is \(\frac{u^{n+1}}{n+1}\).

In our case:

• The integrand \(\sqrt{u}\) is rewritten as \(u^{1/2}\).
• Apply the power rule to get \(\frac{2}{3}u^{3/2}\).
• Finally, evaluate this from 0 to 1, leading to \(3 \times \left( \frac{2}{3}(1) - \frac{2}{3}(0) \right) \).

This cumulative application of substitution and integration techniques results in the solution, confirming that the area under the curve is indeed \(2\) square units.