## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 18 - Fundamental Theorems of Vector Analysis - 18.1 Green's Theorem - Exercises - Page 983: 5

#### Answer

$-\dfrac{\pi}{4}$

#### Work Step by Step

Green's Theorem states that: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$ Here, we have $f=x^2y ; g =0$ Now, $\dfrac{\partial g}{\partial x} =0$ and $\dfrac{\partial f}{\partial y}=x^2$ Therefore, $\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA=\iint_D -x^2 \ dx \ dy$ Now, in polar co-ordinates, we have $\int_C x^2y \ dx=\int_0^{2 \pi}\int_0^{1} -r^2(r) \ d\theta \ dr \\=\int_0^{2 \pi}\int_0^{1} -r^3 \cos^2 \theta \ d\theta \ dr \\=[\int_0^{2 \pi} \cos^2 \theta \ d\theta] \times [-\int_0^{1} r^3 \ dr] \\=[\dfrac{\theta}{2}+\dfrac{\sin 2 \theta}{4}]_0^{2 \pi} \times [\dfrac{-r^4}{4}]_0^1$ Thus, $\int_C x^2y \ dx =-\dfrac{\pi}{4}$

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