Discussion
Domain \(A\) is bounded by curve \(x^{2} = 4y\), ordinate \(x = 2\), and \(x\) axis.
The value of \(\iint_{A} y\, dx\, dy\) is
Answer: A
The integral equals 1/5.
The region lies under the parabola \(y = x^{2}/4\), above the \(x\) axis, for \(0 \le x \le 2\). Integrate \(y\) first:
\[ \begin{aligned} \iint_{A} y\,dx\,dy &= \int_{0}^{2}\int_{0}^{x^{2}/4} y\,dy\,dx \\ &= \int_{0}^{2}\frac{x^{4}}{32}\,dx = \frac{32}{160} = \frac{1}{5} \end{aligned} \]
A common slip is to integrate \(y\) up to 1 or to use the area of the region, \(\int_{0}^{2}x^{2}/4\,dx = 2/3\), in place of the moment; 1/3, 5/12 and 1/2 do not follow from the limits \(0 \le y \le x^{2}/4\).
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