# Chapter 15 - Section 15.7 - Stokes' Theorem - Exercises - Page 896: 7

$-6 \pi$

#### Work Step by Step

Applying Stoke's Theorem, we have $\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$ or, $\oint F \cdot dr=\int_{0}^{2 \pi} \int_{0}^{2} (-3x^2y^2) dy dx$ or, $\int_{0}^{2 \pi} \int_{0}^{2}-6 \sin^2 t+18 \cos^3 t dt=\int_{0}^{2 \pi} -6 \sin^2 t+18 \cos t-18 \cos t\sin^2 t dt$ This implies that $[-3t+3 \sin t \cos t+18 \sin t-6 \sin^3 t]_{0}^{2 \pi}=-6 \pi-0=-6 \pi$

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