Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$x=3 \sin y \sqrt{\cos y} \quad$$ and $$\quad x=0, \quad 0 \leq y \leq \pi / 2$$

Short answer

The area of the region enclosed by the lines and curves is 2 square units.

Step-by-step solution

Establish the bounds and the function to integrate

The integral will range from 0 to \(\pi /2\) according to the given \(y\). To find the function for \(x\), observe that the equation \(x=3 \sin y \sqrt{\cos y}\) is already isolated. Therefore, the function to integrate is \(3 \sin y \sqrt{\cos y}\).

Set up the integral

Simply the line \(x=0\) denotes the y-axis, so the area will be the difference of the curve's x-value and the line's x-value on the interval from 0 to \(\pi /2\), which can be expressed by the following integral: \[\int_0^{\pi /2} (3 \sin y \sqrt{\cos y} - 0) dy\].

Evaluate the integral

Perform the integral operation on \[3 \sin y \sqrt{\cos y}\]. This may require the use of methods like substitution. Here, a substitution \(u=\cos y\) can simplify things since differential \(du=-\sin ydy\). Therefore, the integral becomes \[-3\int_0^1 u^{1/2}du = -3\left[\frac{2}{3}u^{3/2}\right]_0^1 = -3\left[\frac{2}{3}\cdot 1^{3/2} - \frac{2}{3}\cdot0^{3/2}\right]\].

Final calculation and result

The final calculation simplifies to \[-3\cdot\frac{2}{3} = -2\]. We discard negative value because area cannot be negative, so the area of the region is 2 square units.