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You have ﬁfteen minutes to complete the quiz. If anything seems unclear,
please ask.
1. Pacman consists of the space above the region pictured, and below the
upper hemisphere center (0, 0, 0) and radius 1, together with the corresponding space below the xy plane and above the lower hemisphere. Find
the volume of Pacman. (4 points) 2. Set up integrals of a general function f (x, y ) over the regions pictured.
Use whichever of Cartesian and polar coordinates gives a simpler integral.
(3 points) 3. The ﬂoor of a box, modeled by [0, 2] × [−1, 1] is heated, with the heat
coming in at the left hand edge. The position of a mouse is governed by
the probability density function
m(x, y ) = (2 − x)/4
0 if 0 ≤ x ≤ 2 and −1 ≤ y ≤ 1
.
otherwise Find the probability the mouse is in the lefthand (warmer) half of the
box. (3 points) 1 Solutions
If you have any questions and / or think you’ve spotted a mistake / typo, please
get in touch with me.
1. The formula for the hemisphere given in polar coordinates is
1 − r2 . f (r, θ) =
The volume we want is thus equal to
1 7π/4 1 − r2 rdrdθ V =2
0 π/4 (the ‘2’ is there as the integral only gives the ‘upper half’ of Pacman).
Solving this, the inner integral is
1 1 3
1
1 − r2 rdr = − (1 − r2 ) 2
3 0 =
0 1
.
3 The outer integral is
7π/4
π/4 1
6π/4
π
1
dθ = (7π/4 − π/4)) =
=.
3
3
3
2 Hence the volume of Pacman is π .
Alternatively, note that the volume of a sphere of radius 1 is 4π/3, and
that Pacman is 3/4 of a sphere, so has volume (3/4)(4/3)π = π .
2. The ﬁrst one is easier in polar coordinates, and is given by
0 1 f (r cos(θ), r sin(θ))rdrdθ.
−π/4 0 The second one is easier in Cartesian coordinates, and is given by
√ 0 3 /2 0
√
−x/ 3 f (x, y )dydx. 3. The probability we want is
1
−1 1
0 2−x
1
1
dxdy = 2x − x2
4
2
2 2 1 =
0 13
3
=.
22
4 ...
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This note was uploaded on 09/14/2011 for the course MATH 175 taught by Professor Aldroubi during the Spring '08 term at Vanderbilt.
 Spring '08
 ALDROUBI

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