Question

farmer jones, and his wife, dr. jones, decide to build a fence in their field, to keep the sheep safe. since dr. jones is a mathematician, she suggests building fences described by $y = 10x^2$ and $y = x^2 + 11$. farmer jones thinks this would be much harder than just building an enclosure with straight sides, but he wants to please his wife. what is the area of the enclosed region?

Explanation:

Step1: Find intersection x-values

Set $10x^2 = x^2 + 11$
$9x^2 = 11$
$x^2 = \frac{11}{9}$
$x = \pm\frac{\sqrt{11}}{3}$

Step2: Define integrand (upper - lower)

Upper curve: $y = x^2 + 11$, Lower curve: $y = 10x^2$
Integrand: $(x^2 + 11) - 10x^2 = 11 - 9x^2$

Step3: Set up definite integral

Use symmetry, integrate from $0$ to $\frac{\sqrt{11}}{3}$, double the result:
$Area = 2\int_{0}^{\frac{\sqrt{11}}{3}} (11 - 9x^2) dx$

Step4: Compute the integral

Antiderivative: $11x - 3x^3$
Evaluate at bounds:
At $x=\frac{\sqrt{11}}{3}$: $11(\frac{\sqrt{11}}{3}) - 3(\frac{\sqrt{11}}{3})^3 = \frac{11\sqrt{11}}{3} - 3(\frac{11\sqrt{11}}{27}) = \frac{11\sqrt{11}}{3} - \frac{11\sqrt{11}}{9} = \frac{22\sqrt{11}}{9}$
At $x=0$: $0 - 0 = 0$
Multiply by 2: $2\times\frac{22\sqrt{11}}{9} = \frac{44\sqrt{11}}{9}$

Answer:

$\frac{44\sqrt{11}}{9}$