Multiple Integrals Questions and Answers

Answer: B

Integrate the spherical volume element over the full ranges:

\[ V \;=\; \int_{0}^{2\pi}\!\!\int_{0}^{\pi}\!\!\int_{0}^{a} r^{2}\sin\phi\,dr\,d\phi\,d\theta \;=\; (2\pi)(2)\left(\frac{a^{3}}{3}\right) \;=\; \frac{4}{3}\pi a^{3}. \]

The three factors are the longitude range, the integral of \( \sin\phi \) over half a turn, and the radial integral. The value \( 4\pi a^{2} \) is the surface area instead

Answer: A

Each element carries mass \( \rho\,dV \), and summing gives

\[ m \;=\; \iiint \rho(x,y,z)\,dV. \]

Taking \( \rho \) outside the integral is valid only for a uniform solid; for a variable density it must stay inside, which is the whole reason the integral is needed rather than a product

Answer: D

The moment of inertia weights each element by the square of its distance from the axis, and the distance from the \( z \)-axis is \( \sqrt{x^{2}+y^{2}} \), whose square is \( x^{2}+y^{2} \). Including \( z^{2} \) would give the polar moment about the origin, and using the first power of the distance would give a first moment rather than a second

Answer: C

The result is

\[ \frac{\Gamma(l)\Gamma(m)\Gamma(n)}{\Gamma(l+m+n+1)}, \]

and the extra unit in the denominator's argument is what distinguishes it from the beta-function pattern. Setting \( l = m = n = 1 \) gives \( 1/\Gamma(4) = 1/6 \), the volume of the simplex, which is the check that fixes the form

Answer: B

Cartesian coordinates need no factor, all three edges being lengths already. Cylindrical carries one factor of \( r \) for the single angle. Spherical carries \( r^{2}\sin\phi \), one power of the radius for each of its two angles and a sine because circles of latitude shrink towards the poles

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