Discussion
Written the other way round, \( \displaystyle\int_{0}^{1}\!\!\int_{0}^{\sqrt{x}} f\,dy\,dx \) carries inner limits for \( x \) running from
Answer: B
The region lies below \( y = \sqrt{x} \), that is above \( x = y^{2} \), with \( y \) running from zero to one. So for a fixed \( y \) the variable \( x \) runs from the curve across to the boundary:
\[ \int_{0}^{1}\!\!\int_{y^{2}}^{1} f\,dx\,dy. \]
Taking \( x \) from zero to \( y^{2} \) describes the region on the *other* side of the same curve
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