Multiple Integrals Questions and Answers

Answer: A

The integral equals \(1 - \cos(1)\).

The limits are constants, so the order of integration can be swapped. Integrating with respect to \(y\) first is easy, because \(x\) is exactly the factor produced by differentiating \(\sin(xy)\) with respect to \(y\):

\[ \begin{aligned} \int_{0}^{1} x\cos(xy)\, dy &= \Big[ \sin(xy) \Big]_{y=0}^{y=1} = \sin x \\ I &= \int_{0}^{1} \sin x\, dx \\ &= \Big[ -\cos x \Big]_{0}^{1} = 1 - \cos 1 \end{aligned} \]

Numerically this is about 0.46. \(1 + \cos 1\) has the sign of the cosine term wrong, and the two options with \(\sin 1\) come from integrating \(\cos\) instead of \(\sin\) in the last step. Integrating in the written order (first in \(x\)) needs integration by parts and gives the same result with more work.

Answer: A

The integral equals \(\pi(1 - e^{-4})\).

Use polar coordinates, \(x^{2} + y^{2} = r^{2}\) and \(dx\,dy = r\,dr\,d\theta\), with \(0 \le r \le 2\) and \(0 \le \theta \le 2\pi\):

\[ \begin{aligned} I &= \int_{0}^{2\pi}\int_{0}^{2} e^{-r^{2}} r\,dr\,d\theta \\ &= 2\pi \left[-\frac{1}{2}e^{-r^{2}}\right]_{0}^{2} \\ &= 2\pi \cdot \frac{1}{2}(1 - e^{-4}) = \pi(1 - e^{-4}) \end{aligned} \]

The exponent is \(-r^{2} = -4\) at the edge of the disc of radius 2, so the options with \(e^{-2}\) use the radius instead of its square. The options with \(\frac{\pi}{2}\) lose the \(2\pi\) from the angle, or integrate only over a half turn.

Answer: B

The answer is \(\int_0^1 \int_{x^2}^{x^2+1} dy\, dx\).

Read off the region. In the first integral \(0 \le y \le 1\) and \(0 \le x \le \sqrt{y}\), that is \(y \ge x^2\). In the second, \(1 \le y \le 2\) and \(\sqrt{y - 1} \le x \le 1\), that is \(y \le x^2 + 1\). Together they cover, for each x in [0,1], the strip between the two parabolas:

\[ 0 \le x \le 1, \quad x^2 \le y \le x^2 + 1 \]

so a single integral in the order dy dx suffices. The option with upper limit \(x^2 - 1\) has the limits the wrong way round (it gives a negative area), and the two split options use boundaries such as \(y = \pm x\) and \(y = \sqrt{x}\) that are not edges of this region.

Answer: A

The integral equals 1/5.

The region lies under the parabola \(y = x^{2}/4\), above the \(x\) axis, for \(0 \le x \le 2\). Integrate \(y\) first:

\[ \begin{aligned} \iint_{A} y\,dx\,dy &= \int_{0}^{2}\int_{0}^{x^{2}/4} y\,dy\,dx \\ &= \int_{0}^{2}\frac{x^{4}}{32}\,dx = \frac{32}{160} = \frac{1}{5} \end{aligned} \]

A common slip is to integrate \(y\) up to 1 or to use the area of the region, \(\int_{0}^{2}x^{2}/4\,dx = 2/3\), in place of the moment; 1/3, 5/12 and 1/2 do not follow from the limits \(0 \le y \le x^{2}/4\).

Answer: A

The surface area is \(\frac{\pi}{6}(2\sqrt{2} - 1)\).

For a surface \(z = g(x,y)\), the area element is \(\sqrt{1 + g_{x}^{2} + g_{y}^{2}}\,dA = \sqrt{1 + 4r^{2}}\,dA\). The portion below \(z = \frac{1}{4}\) lies over the disc \(r \le \frac{1}{2}\). In polar coordinates:

\[ \begin{aligned} S &= \int_{0}^{2\pi}\!\int_{0}^{1/2} \sqrt{1 + 4r^{2}}\, r\,dr\,d\theta \\ &= 2\pi \left[ \tfrac{1}{12}(1 + 4r^{2})^{3/2} \right]_{0}^{1/2} \\ &= \tfrac{\pi}{6}\left(2^{3/2} - 1\right) = \tfrac{\pi}{6}(2\sqrt{2} - 1) \end{aligned} \]

The other options carry the same bracket with a wrong factor: \(\frac{\pi}{2}\), \(\frac{\pi}{3}\) and \(\pi\) come from losing the \(\frac{1}{12}\) of the substitution \(u = 1 + 4r^{2}\) or the \(\frac{1}{2}\) radius limit.

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