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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Written the other way round, \( \displaystyle\int_{0}^{1}\!\!\int_{0}^{\sqrt{x}} f\,dy\,dx \) carries inner limits for \( x \) running from
Answer: B
The region lies below \( y = \sqrt{x} \), that is above \( x = y^{2} \), with \( y \) running from zero to one. So for a fixed \( y \) the variable \( x \) runs from the curve across to the boundary:
\[ \int_{0}^{1}\!\!\int_{y^{2}}^{1} f\,dx\,dy. \]
Taking \( x \) from zero to \( y^{2} \) describes the region on the *other* side of the same curve
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The average value of \( f(x,y) \) over a region \( R \) of area \( A \) is
Answer: A
As in one dimension, the average is the total divided by the extent over which it is spread:
\[ \bar{f} \;=\; \frac{1}{A}\iint_{R} f\,dA, \qquad A \;=\; \iint_{R} dA. \]
A constant integrand then returns its own value, which is the test any averaging formula must pass. The version with \( f^{2} \) leads to a root-mean-square instead
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Match the region with the natural coordinate system for integrating over it. 1. A rectangle 2. A disc 3. A ball P. Plane polar Q. Cartesian R. Spherical
Answer: D
The boundary of the region should become a coordinate held constant. A rectangle has sides of constant \( x \) or \( y \); a disc has a boundary of constant \( r \) in the plane; a ball has a boundary of constant \( r \) in space, which is the spherical radius. Matching the coordinates to the boundary is what turns variable limits into constant ones
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Put these steps of finding a volume by triple integration in order. 1. Integrate in the innermost variable 2. Choose the coordinate system that makes the boundary a constant coordinate 3. Work outwards through the remaining two integrals 4. Write the three ranges, innermost first
Answer: C
The coordinate system is chosen first, because it decides whether the limits are constants or functions. The three ranges are then written from the inside out, and the integrals evaluated in that order, each result becoming the integrand of the next. Choosing coordinates after writing the limits means writing them twice
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None of these four integrals gives an area except
Answer: B
Only the unit integrand accumulates area itself. Weighting by \( x \) gives a first moment, whose dimensions are a length cubed; weighting by \( x^{2}+y^{2} \) gives a second moment; and the triple integral of unity gives a volume. Each of those three answers a different question about the same region
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