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\begin{center}
\textbf{Multiple Integration}

\textit{\textbf{Double Integrals}}
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\textbf{Question}

Evaluate the following double integral by inspection.

$\int \! \int_D (x+3) \,dA$,

where $D$ is the half-disk $o \le y \le \sqrt{4-x^2}$.


\textbf{Answer}

$\begin{array}{ll}
\int \! \int_D (x+3 \,dA  & = \int \! \int_D x \,dA +3\int \! \int_D
\,dA \\
& = 0 + 3(\textrm{area of }D)\\
& = 3 \times \frac{\pi 2^2}{2} = 6\pi
\end{array}
\ \ \
\begin{array}{c}
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\end{array}$

The integral of $x$ over $D$ equals zero because $D$ is symmetrical
about $x=0$. 


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