Problem 33 - 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 in the first quadrant bounded by \(y=x^{2 / 3}\) and \(y=4\)

Step-by-Step Solution

Answer

Answer: The area of the region is \(12.8\) square units.

Step 1: Sketch the Bounding Curves and Region

First, let's sketch the two given functions and the region in question. The first function, \(y = x^{2/3}\), is a curve that increases at a decreasing rate as x increases. The second function, \(y = 4\), is simply a horizontal line. Plot the functions and shade the region enclosed by the curves and bounded by the first quadrant. Notice that the curve \(y = x^{2/3}\) intersects the line \(y = 4\) at two points: \(x = 0\) and another x-coordinate we need to find.

Step 2: Find Points of Intersection

To find the points of intersection of the functions \(y=x^{2/3}\) and \(y=4\), we need to set the equations equal to each other: \(x^{2/3} = 4\) Now, we'll solve for x: Raise both sides of the equation to the power of 3 to cancel out the \(x^{2/3}\) term: \(x^2 = 4^3\) \(x^2 = 64\) Now, take the square root of both sides: \(x = \pm 8\) Since the question asks for the region in the first quadrant, we're only concerned with the positive x-value: \(x = 8\) Thus, the points of intersection occur at \((0, 4)\) and \((8, 4)\).

Step 3: Set up Integral to Find Area

To find the area of the region enclosed by the curves and in the first quadrant, we'll integrate the difference of the functions with respect to x across the interval \([0, 8]\): \(\displaystyle\int_\limits{0}^{8} (4 - x^{2/3}) dx\)

Step 4: Calculate the Integral

Now, we'll evaluate the integral to find the area of the region: \( \displaystyle\int_\limits{0}^{8} (4 - x^{2/3}) dx = \left[4x - \frac{3}{5}x^{5/3} \right]_0^8 \) Plug in the limits of integration: \(\left(4(8) - \frac{3}{5}(8)^{5/3}\right) - \left(4(0) - \frac{3}{5}(0)^{5/3}\right)\) Simplify and compute: \((32 - \frac{3}{5}(32)) - (0) = 32 - 19.2 = 12.8\) Therefore, the area of the region in the first quadrant bounded by \(y = x^{2/3}\) and \(y = 4\) is \(12.8\) square units.