Multiple Integrals Questions and Answers

Answer: C

Each element of the base carries a column of height \( f \) and base area \( dA \), so its volume is \( f\,dA \) and

\[ V \;=\; \iint_{R} f(x,y)\,dA. \]

The squared integrand with \( \pi \) belongs to a solid of revolution, whose slices are discs rather than columns, and the square root form gives a surface area

Answer: B

Integrate the height \( z = 1 - x - y \) over the triangle \( x + y \le 1 \) in the base plane:

\[ V \;=\; \int_{0}^{1}\!\!\int_{0}^{1-x}\left(1 - x - y\right)dy\,dx \;=\; \int_{0}^{1}\frac{(1-x)^{2}}{2}\,dx \;=\; \frac{1}{6}. \]

The elementary formula \( \tfrac{1}{3}\times\text{base}\times\text{height} = \tfrac{1}{3}(\tfrac{1}{2})(1) \) agrees

Answer: A

The curves meet where \( x^{2} = 2x \), that is at \( x = 0 \) and \( x = 2 \), with the line above the parabola between them:

\[ A \;=\; \int_{0}^{2}\!\!\int_{x^{2}}^{2x} dy\,dx \;=\; \int_{0}^{2}\left(2x - x^{2}\right)dx \;=\; \left[x^{2} - \frac{x^{3}}{3}\right]_{0}^{2} \;=\; 4 - \frac{8}{3} \;=\; \frac{4}{3}. \]

The inner integral returns the vertical extent of the region at each \( x \), so the two areas four and \( 8/3 \) are the separate contributions rather than the answer

Answer: D

Cylindrical coordinates are plane polars with \( z \) carried along unchanged, so the plane element \( r\,dr\,d\theta \) is simply multiplied by the height:

\[ dV \;=\; r\,dr\,d\theta\,dz. \]

The form with \( r^{2}\sin\phi \) is the spherical element, where both angles are true angles and two factors of length are needed to convert them

Answer: C

The three edges of a small box are \( dr \) radially, \( r\,d\phi \) along the meridian and \( r\sin\phi\,d\theta \) around the axis, the last shrinking towards the poles. Their product is

\[ dV \;=\; r^{2}\sin\phi\,dr\,d\phi\,d\theta. \]

The sine factor is what makes the element vanish on the axis, where a change of longitude moves no distance at all

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