Three of these four hold for the double integral. The exception is
Answer: D
For a continuous integrand over a reasonable region the value is the same in either order, which is Fubini's theorem; what changes is the difficulty. Reduction to a repeated integral, invariance under a properly weighted change of variables, and the area interpretation are all correct
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True or false: the limits of a double integral may be interchanged along with the differentials, leaving everything else alone.
Answer: C
Only for a rectangle with constant limits does the swap leave the numbers unchanged. In general the limits describe the region in one particular order, and reversing the order means re-describing the same region, which needs the sketch. Continuity guarantees the two orders give the same *value*, not the same limits
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True or false: the Jacobian factor in a change of variables may be negative.
Answer: B
The determinant itself may well be negative, and its sign does record a reversal of orientation. But the area element uses its modulus, so the factor entering the integral is never negative; otherwise a positive integrand over a genuine region could return a negative total
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Evaluating \( \displaystyle\int_{0}^{1}\!\!\int_{0}^{1}\left(x + y\right)dy\,dx \) gives
Answer: A
Do the inner integral with \( x \) held fixed:
\[ \int_{0}^{1}\left(x + y\right)dy \;=\; x + \frac{1}{2}. \]
Then the outer one:
\[ \int_{0}^{1}\left(x + \frac{1}{2}\right)dx \;=\; \frac{1}{2} + \frac{1}{2} \;=\; 1. \]
By symmetry each variable contributes \( 1/2 \), so the halves add rather than averaging to one of them
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Integrating \( xy \) over the triangle bounded by \( x = 0 \), \( y = 0 \) and \( x + y = 1 \) yields
Answer: D
For each \( x \) the variable \( y \) runs up to \( 1 - x \):
\[ \int_{0}^{1}\!\!\int_{0}^{1-x} xy\,dy\,dx \;=\; \int_{0}^{1} x\,\frac{(1-x)^{2}}{2}\,dx. \]
Expanding gives \( \tfrac{1}{2}\int_{0}^{1}\left(x - 2x^{2} + x^{3}\right)dx = \tfrac{1}{2}\left(\tfrac{1}{2} - \tfrac{2}{3} + \tfrac{1}{4}\right) = \tfrac{1}{24} \). Over the whole unit square the same integrand would give \( 1/4 \)
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The double integral of unity over the disc \( x^{2} + y^{2} \le 9 \) equals
Answer: C
An integrand of one returns the area, and in polar form
\[ \int_{0}^{2\pi}\!\!\int_{0}^{3} r\,dr\,d\theta \;=\; 2\pi\left[\frac{r^{2}}{2}\right]_{0}^{3} \;=\; 2\pi\left(\frac{9}{2}\right) \;=\; 9\pi. \]
This is \( \pi r^{2} \) with \( r = 3 \). Taking the radius as nine rather than its square would give \( 81\pi \)
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Integrating \( r \) over the disc \( r \le a \) in polar coordinates gives
Answer: B
The integrand is \( r \) and the element supplies another factor of \( r \):
\[ \int_{0}^{2\pi}\!\!\int_{0}^{a} r \cdot r\,dr\,d\theta \;=\; 2\pi\left[\frac{r^{3}}{3}\right]_{0}^{a} \;=\; \frac{2\pi a^{3}}{3}. \]
Two powers of \( r \) appear, one from the integrand and one from the area element, which is what raises the result to a cube
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The volume of the cylinder \( x^{2} + y^{2} \le 4 \) between \( z = 0 \) and \( z = 5 \) by triple integration is
Answer: A
In cylindrical coordinates the three integrals separate:
\[ V \;=\; \int_{0}^{2\pi}\!\!\int_{0}^{2}\!\!\int_{0}^{5} r\,dz\,dr\,d\theta \;=\; (2\pi)\left(\frac{4}{2}\right)(5) \;=\; 20\pi. \]
This is \( \pi r^{2}h = \pi(4)(5) \). Using the radius rather than its square would double the answer to \( 40\pi \)
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The centroid of the quarter disc \( x^{2}+y^{2} \le a^{2} \) with \( x, y \ge 0 \) has \( \bar{x} = \)
Answer: D
Divide the first moment by the area, working in polar coordinates with \( x = r\cos\theta \):
\[ \iint x\,dA \;=\; \int_{0}^{\pi/2}\!\!\int_{0}^{a} r\cos\theta \cdot r\,dr\,d\theta \;=\; \frac{a^{3}}{3}. \]
The area is \( \pi a^{2}/4 \), so
\[ \bar{x} \;=\; \frac{a^{3}/3}{\pi a^{2}/4} \;=\; \frac{4a}{3\pi} \;\approx\; 0.424a. \]
By symmetry \( \bar{y} \) takes the same value
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A uniform disc of radius \( a \) and surface density \( \sigma \) has, about its own axis, a second moment of
Answer: C
Weight each element by the square of its distance from the axis:
\[ I \;=\; \sigma\int_{0}^{2\pi}\!\!\int_{0}^{a} r^{2}\cdot r\,dr\,d\theta \;=\; 2\pi\sigma\left[\frac{r^{4}}{4}\right]_{0}^{a} \;=\; \frac{\pi\sigma a^{4}}{2}. \]
Since the mass is \( m = \pi\sigma a^{2} \), this is \( \tfrac{1}{2}ma^{2} \), the familiar disc result
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