Problem 37 - Calculus Early Transcendentals

Question

Use any method (including geometry) to find the area of the following regions. In each case, sketch the bounding curves and the region in question. The region between the line \(y=x\) and the curve \(y=2 x \sqrt{1-x^{2}}\) in the first quadrant

Step-by-Step Solution

Answer

Answer: The area of the region bounded by the curves in the first quadrant is \(\frac{1}{48}\).

Step 1: Identify the Points of Intersection

We start by setting the two functions equal to each other to find the points of intersection in the first quadrant: $$x = 2x\sqrt{1-x^{2}}$$ First, notice that the point \((0,0)\) is a point of intersection, as both functions are equal to zero at this point. Next, we can divide by \(x\), seeing that we already considered the solution \(x=0\): $$1 = 2\sqrt{1-x^{2}}$$ Now, we can continue to isolate \(x^2\): $$\frac{1}{2} = \sqrt{1-x^2} \Rightarrow \frac{1}{4}=1-x^2 \Rightarrow x^2=\frac{3}{4} \Rightarrow x=\frac{\sqrt{3}}{2}$$ So, the points of intersection are \((0,0)\) and \((\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2})\).

Step 2: Set up the Integral

We want to calculate the area of the region bounded by the two curves in the first quadrant. To do this, we'll use integration. Since we're dealing with a region bounded on the left and right, it's more natural to integrate with respect to \(x\). We set up the integral like so: $$A = \int_{0}^{\frac{\sqrt{3}}{2}} [2x\sqrt{1-x^{2}} - x] \, dx$$ This integral represents the difference between the functions \(2x\sqrt{1-x^{2}}\) and \(x\) over the interval between the intersection points.

Step 3: Evaluate the Integral

Now, we'll evaluate the integral: $$A = \int_{0}^{\frac{\sqrt{3}}{2}} [2x\sqrt{1-x^{2}} - x] \, dx$$ To handle the first term of the integral \(2x\sqrt{1-x^2}\), we can use substitution. Let \(u=1-x^2\), so \(-2x\,dx=du\). Our integral now becomes: $$A = \int_{1}^{\frac{1}{4}} [\sqrt{u} - \frac{u}{2}] \, du$$ $$A = \frac{2}{3}u^{\frac{3}{2}} - \frac{1}{4}u^2\Bigg|_{1}^{\frac{1}{4}}$$ After plugging in the bounds of integration: $$A = \left(\frac{2}{3}\left(\frac{1}{4}\right)^{\frac{3}{2}} - \frac{1}{16}\right) - \left(\frac{2}{3} - \frac{1}{4}\right)$$ $$A = \frac{2}{3}\left(\frac{1}{8}\right) - \frac{1}{16} - \frac{2}{3} + \frac{1}{4}$$ Finally, simplify and find the area: $$A = \frac{1}{12} - \frac{1}{16} - \frac{2}{3} + \frac{1}{4} = \boxed{\frac{1}{48}}$$ Thus, the area of the region bounded by the two curves in the first quadrant is \(\frac{1}{48}\).