Chez lefouque 乐福家园: A Polar Double Integral

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Tuesday, April 12, 2011

A Polar Double Integral

$\int_{0}^{\frac{π}{2}} \int_{0}^{1 + sin θ} θ r d r d θ$
=

$\int_{0}^{\frac{π}{2}} θ {[\frac{1}{2} r^{2}]}_{0}^{1 + sin θ} d θ$
=

$\int_{0}^{\frac{π}{2}} \frac{1}{2} θ {(1 + sin θ)}^{2} d θ$
=

$\frac{1}{2} \int_{0}^{\frac{π}{2}} θ (1 + 2 sin θ + {sin}^{2} θ) d θ$
=

$\frac{1}{2} \int_{0}^{\frac{π}{2}} θ (1 + 2 sin θ + \frac{1 - cos 2 θ}{2}) d θ$
=

$\frac{1}{2} \int_{0}^{\frac{π}{2}} θ (\frac{3}{2} + 2 sin θ - \frac{1}{2} cos 2 θ) d θ$

=

$\frac{3}{4} \int_{0}^{\frac{π}{2}} θ d θ + \int_{0}^{\frac{π}{2}} θ (sin θ - \frac{1}{4} cos 2 θ) d θ$
=

$\frac{3}{4} {[\frac{θ^{2}}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} θ \frac{d}{d θ} (cos θ + \frac{1}{8} sin 2 θ) d θ$
=

$\frac{3}{8} \cdot \frac{π^{2}}{4} - {[θ (cos θ + \frac{1}{8} sin 2 θ)]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} cos θ + \frac{1}{8} sin 2 θ d θ$
=

$\frac{3 π^{2}}{32} - 0 + {[sin θ - \frac{1}{16} cos 2 θ]}_{0}^{\frac{π}{2}}$
=

$\frac{3 π^{2}}{32} + [(1 + \frac{1}{16}) - (0 - \frac{1}{16})]$
=

$\frac{3 π^{2}}{32} + \frac{9}{8}$

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