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.