Answer
\[ \oint_C \left( x^2 + y^2 \right) \, ds = \frac{1}{30} \oint_C \left( e^x + y^2 \, dx + e^y + x^2 \, dy \right) = \frac{1}{30} \]
Work Step by Step
Using Green's theorem: Let \( \Phi(x, y) = x^2 + y^2 \) and \( \Psi(x, y) = x^2 + y^2 \), and \( C \) is the given curve. \[ \oint_C \left( \Phi dx + \Psi dy \right) = \iint_R \left( \frac{\partial \Psi}{\partial x} - \frac{\partial \Phi}{\partial y} \right) dA \] Where \( f(x, y) = e^x + y^2 \) and \( g(x, y) = e^y + x^2 \), and \( C \) is the given curve. \[ \oint_C \left( e^x + y^2 \, dx + e^y + x^2 \, dy \right) = \iint_R \left( \frac{\partial}{\partial x} \left( e^y + x^2 \right) - \frac{\partial}{\partial y} \left( e^x + y^2 \right) \right) dA \] \[ = \int_0^1 \int_{x^2}^{x} 2(x - y) \, dy \, dx \] \[ = \int_0^1 \left[ 2x^2 - 2x^3 - x^2 + x^4 \right] \, dx \] \[ = \int_0^1 \left( x^4 - 2x^3 + x^2 \right) \, dx \] \[ = \left. \frac{1}{5}x^5 - \frac{1}{2}x^4 + \frac{1}{3}x^3 \right|_0^1 \] \[ = \frac{1}{5} - \frac{1}{2} + \frac{1}{3} \] \[ = \frac{6}{30} - \frac{15}{30} + \frac{10}{30} \] \[ = \frac{1}{30} \] Finally, the result is: \[ \oint_C \left( x^2 + y^2 \right) \, ds = \frac{1}{30} \oint_C \left( e^x + y^2 \, dx + e^y + x^2 \, dy \right) = \frac{1}{30} \]
