Area Between Curves: $y=\sqrt{x}$ and $y=8x^2$

This problem involves finding the area enclosed by two specific curves, $y = \sqrt{x}$ and $y = 8x^2$. We need to find where they intersect and then calculate the definite integral of the difference between the upper and lower curves.

Intersection Points Identification

First, find the points where the curves intersect by setting their equations equal:

$ \sqrt{x} = 8x^2 $

Square both sides to eliminate the square root:

$ x = (8x^2)^2 $

$ x = 64x^4 $

Rearrange into a polynomial equation:

$ 64x^4 - x = 0 $

Factor out $x$:

$ x(64x^3 - 1) = 0 $

The solutions are $x=0$ and $64x^3 - 1 = 0$. Solving the second equation gives:

$ 64x^3 = 1 \implies x^3 = \frac{1}{64} \implies x = \frac{1}{4} $

The intersection points occur at $x=0$ and $x=1/4$. These serve as the limits for our integral.

Upper Function Determination

We need to determine which function is greater (the upper curve) within the interval $[0, 1/4]$. Choose a test point, like $x=1/16$:

• $y = \sqrt{1/16} = 1/4$
• $y = 8(1/16)^2 = 8(1/256) = 1/32$

Since $1/4 > 1/32$, $y = \sqrt{x}$ is the upper curve in this interval.

Integral Evaluation for Area

The area $A$ is the definite integral of the upper curve minus the lower curve from $x=0$ to $x=1/4$:

$ A = \int_{0}^{1/4} (\sqrt{x} - 8x^2) dx $

Rewrite $\sqrt{x}$ as $x^{1/2}$ and integrate:

$ A = \int_{0}^{1/4} (x^{1/2} - 8x^2) dx $

$ A = \left[ \frac{x^{3/2}}{3/2} - 8\frac{x^3}{3} \right]_{0}^{1/4} $

$ A = \left[ \frac{2}{3}x^{3/2} - \frac{8}{3}x^3 \right]_{0}^{1/4} $

Evaluate using the limits:

$ A = \left( \frac{2}{3}\left(\frac{1}{4}\right)^{3/2} - \frac{8}{3}\left(\frac{1}{4}\right)^3 \right) - \left( \frac{2}{3}(0)^{3/2} - \frac{8}{3}(0)^3 \right) $

$ A = \left( \frac{2}{3} \cdot \frac{1}{8} - \frac{8}{3} \cdot \frac{1}{64} \right) - 0 $

$ A = \frac{2}{24} - \frac{8}{192} $

$ A = \frac{1}{12} - \frac{1}{24} $

$ A = \frac{2}{24} - \frac{1}{24} = \frac{1}{24} $

Final Area Rounding

Convert the fractional result to a decimal:

$ A = \frac{1}{24} \approx 0.041666... $

Rounding to three decimal places, the area is $0.042$. This value is consistent with the provided answer range (0.03 to 0.05).