Multiple Integrals Questions and Answers

Answer: D

A small patch spans \( dr \) radially and \( r\,d\theta \) around, since arc length is radius times angle, so its area is the product

\[ dA \;=\; r\,dr\,d\theta. \]

The factor \( r \) is the Jacobian of the transformation from Cartesian coordinates, and omitting it is the commonest error in polar integration: patches far from the origin would then count for no more than those near it

Answer: C

Integrate the polar area element over the whole disc:

\[ A \;=\; \int_{0}^{2\pi}\!\!\int_{0}^{a} r\,dr\,d\theta \;=\; \int_{0}^{2\pi}\frac{a^{2}}{2}\,d\theta \;=\; \frac{a^{2}}{2}(2\pi) \;=\; \pi a^{2}. \]

Dropping the factor \( r \) would give \( 2\pi a \), which is the circumference and not an area at all

Answer: B

In polar form the exponent becomes \( -r^{2} \) and the area element supplies the factor \( r \) that the substitution \( u = r^{2} \) needs:

\[ \int_{0}^{2\pi}\!\!\int_{0}^{\infty} e^{-r^{2}}r\,dr\,d\theta \;=\; 2\pi \cdot \frac{1}{2} \;=\; \pi. \]

This is the standard route to \( \int_{-\infty}^{\infty} e^{-x^{2}}dx = \sqrt{\pi} \), which has no elementary antiderivative in one variable

Answer: A

The rule is

\[ dx\,dy \;=\; \left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv, \]

the Jacobian of the *old* variables with respect to the new ones, and its modulus because an area is positive whichever way the map orients the plane. Using the reciprocal Jacobian scales the integral by the wrong factor

Answer: D

Taking the integrand as one makes each element contribute its own area, so the total is the area itself:

\[ A \;=\; \iint_{R} dA. \]

An integrand of \( x \) or \( y \) gives a first moment instead, from which a centroid is found, and a product \( xy \) gives a product moment

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