Multiple Integrals Questions and Answers

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

Answer: A

As in one dimension, the average is the total divided by the extent over which it is spread:

\[ \bar{f} \;=\; \frac{1}{A}\iint_{R} f\,dA, \qquad A \;=\; \iint_{R} dA. \]

A constant integrand then returns its own value, which is the test any averaging formula must pass. The version with \( f^{2} \) leads to a root-mean-square instead

Answer: D

The boundary of the region should become a coordinate held constant. A rectangle has sides of constant \( x \) or \( y \); a disc has a boundary of constant \( r \) in the plane; a ball has a boundary of constant \( r \) in space, which is the spherical radius. Matching the coordinates to the boundary is what turns variable limits into constant ones

Answer: C

The coordinate system is chosen first, because it decides whether the limits are constants or functions. The three ranges are then written from the inside out, and the integrals evaluated in that order, each result becoming the integrand of the next. Choosing coordinates after writing the limits means writing them twice

Answer: B

Only the unit integrand accumulates area itself. Weighting by \( x \) gives a first moment, whose dimensions are a length cubed; weighting by \( x^{2}+y^{2} \) gives a second moment; and the triple integral of unity gives a volume. Each of those three answers a different question about the same region