Multiple Integrals Questions and Answers

Answer: A

An integrand of one accumulates area. Weighting by the distance \( x \) gives a first moment, from which the centroid follows on dividing by the area. Weighting by the square gives a second moment, the quantity that appears in bending and in moments of inertia

Answer: D

The sketch has to come first, since the given limits describe the region only implicitly and the new ones cannot be read off without it. The outer variable is then chosen, its full range read from the sketch, and the inner limits expressed in terms of it. Writing the integral is the last step

Answer: C

The first and third are standard. The second is false in general: the limits describe the same region read a different way, and only for a rectangle with constant limits do they survive the exchange unchanged. For a triangle or a region under a curve they must be recomputed from the sketch

Answer: B

The first two hold, and simplifying the region is often the whole reason for the change. The third is false: a linear map has a *constant* Jacobian, which is easy to compute but is not one unless the map preserves area. Scaling by two in each direction multiplies areas by four, and that factor must appear

Answer: A

A wedge of fixed angular width is wider the further out it lies, its arc being \( r\,d\theta \), so a patch far from the origin covers more area than one near it. The factor \( r \) records that. Radians make the arc formula simple but are not the reason, and the behaviour at the origin is a consequence rather than a cause

Answer: D

For a continuous integrand over a reasonable region the value is the same in either order, which is Fubini's theorem; what changes is the difficulty. Reduction to a repeated integral, invariance under a properly weighted change of variables, and the area interpretation are all correct

Answer: C

Only for a rectangle with constant limits does the swap leave the numbers unchanged. In general the limits describe the region in one particular order, and reversing the order means re-describing the same region, which needs the sketch. Continuity guarantees the two orders give the same *value*, not the same limits

Answer: B

The determinant itself may well be negative, and its sign does record a reversal of orientation. But the area element uses its modulus, so the factor entering the integral is never negative; otherwise a positive integrand over a genuine region could return a negative total

Answer: A

Do the inner integral with \( x \) held fixed:

\[ \int_{0}^{1}\left(x + y\right)dy \;=\; x + \frac{1}{2}. \]

Then the outer one:

\[ \int_{0}^{1}\left(x + \frac{1}{2}\right)dx \;=\; \frac{1}{2} + \frac{1}{2} \;=\; 1. \]

By symmetry each variable contributes \( 1/2 \), so the halves add rather than averaging to one of them

Answer: D

For each \( x \) the variable \( y \) runs up to \( 1 - x \):

\[ \int_{0}^{1}\!\!\int_{0}^{1-x} xy\,dy\,dx \;=\; \int_{0}^{1} x\,\frac{(1-x)^{2}}{2}\,dx. \]

Expanding gives \( \tfrac{1}{2}\int_{0}^{1}\left(x - 2x^{2} + x^{3}\right)dx = \tfrac{1}{2}\left(\tfrac{1}{2} - \tfrac{2}{3} + \tfrac{1}{4}\right) = \tfrac{1}{24} \). Over the whole unit square the same integrand would give \( 1/4 \)