Multiple Integrals Questions and Answers

Answer: A

The region is divided into small pieces, the function sampled in each, and \( \sum f(x_{i},y_{i})\,\Delta A_{i} \) formed; the limit as every piece shrinks is the integral. It factors into a product of single integrals only when the region is a rectangle and the integrand itself separates, which is a special case rather than the definition

Answer: D

Integrate inside first, treating \( x \) as constant:

\[ \int_{0}^{2} xy\,dy \;=\; x\left[\frac{y^{2}}{2}\right]_{0}^{2} \;=\; 2x. \]

Then integrate outside:

\[ \int_{0}^{1} 2x\,dx \;=\; \left[x^{2}\right]_{0}^{1} \;=\; 1. \]

The limits are constant and the integrand separates, so the product \( \left(\int_{0}^{1}x\,dx\right)\left(\int_{0}^{2}y\,dy\right) = (1/2)(2) \) gives the same value

Answer: C

When the limits do not depend on the other variable and the integrand factorises, the inner integral of \( g \) is a constant that passes out of the outer one:

\[ \iint f(x)g(y)\,dA \;=\; \left(\int f(x)\,dx\right)\left(\int g(y)\,dy\right). \]

Both conditions are needed: a variable limit or a non-separable integrand destroys the factorisation

Answer: B

The region is the triangle with \( 0 \le x \le 1 \) and \( x \le y \le 1 \), that is the part of the unit square above the line \( y = x \). Describing it the other way, \( y \) runs from zero to one and for each \( y \) the variable \( x \) runs from zero up to \( y \):

\[ \int_{0}^{1}\!\!\int_{0}^{y} f\,dx\,dy. \]

Sketching the region is the only reliable way to get this right

Answer: A

The classic case is \( \int_{0}^{1}\!\int_{x}^{1} e^{-y^{2}}dy\,dx \), where the inner integral cannot be done in the order given but becomes elementary when reversed. With a rectangle and constant limits the order makes no difference to the difficulty, so there is nothing to be gained by swapping

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