Question

The smallar area (in square units) enclosed between the curves $C_1: x^2 + y^2 = 1$ and $C_2: x^{1/3} + y^{1/3} = 1$ is:

• A. $\frac{\pi}{2} - \frac{1}{20}$
• B. $\frac{\pi}{4} - \frac{1}{20}$
• C. $\frac{\pi}{4} - \frac{1}{10}$
• D. $\frac{\pi}{2} - \frac{1}{10}$

Answer 1

Answer: (B)

$\frac{\pi}{4} - \frac{1}{20}$

Answer 2

The problem asks for the smaller area enclosed between a circle $C_1: x^2+y^2=1$ and a curve $C_2: x^{1/3}+y^{1/3}=1$. Both curves pass through (1,0) and (0,1). By analyzing the functions in the first quadrant, it's found that $C_2$ lies inside $C_1$. Therefore, the smaller area is the region in the first quadrant bounded by the two curves. This area is calculated by subtracting the area under $C_2$ from the area under $C_1$ from $x=0$ to $x=1$.

The area under $C_1$ in the first quadrant is a quarter circle, $\frac{\pi}{4}$. The area under $C_2$ in the first quadrant is $\int_0^1 (1-x^{1/3})^3 dx$. Using the substitution $x=t^3$, this integral evaluates to $\frac{1}{20}$. The difference, $\frac{\pi}{4} - \frac{1}{20}$, is the required smaller area.