Consider the 2D Pac-Man, which is a circle of radius 1 with a quarter wedge remo
ID: 2887129 • Letter: C
Question
Consider the 2D Pac-Man, which is a circle of radius 1 with a quarter wedge removed: In the mid nineties Pac-Man evolved to 3D. The game designers decided they wanted the side profile of Pac-Man to remain the same in 2D and 3D. They therefore opted for a sphere with a wedge cut out of it for the mouth. (a) Describe this new region. (b) Using an appropriate integration, determine the volume of Pac-Man (again, you may like to check this with your high school level knowledge regarding geometry of a sphere) (c) What is the surface area of Pac-Man?Explanation / Answer
a> the Pac-Man has a circular region .
We are given that this circle has a redius of 1 units and lets say that the center of this circle is at the orign that is (0,0)
=> The equation of this circular region is : x^2 + y^2 = 1
lets convert this in the polar coordinate system :
In polar coordinate system :
x = rcos(t) , y = rsin(t) and x^2 + y^2 = r^2
=> Our equation becomes : (rcos(t))^2 + (rsin(t))^2 = 1 = r^2
=> r = 1 and t E [pi/4 , 7pi/4] , -------> this is the region for the pac-man
t moves from pi/4 to 7pi/4 as we took out a quarter wedge as show in the figure.
b> The volume is , V = Integral (t = pi/4 to 7pi/4 )Integral(r=0 to 1) rdrdt
= Integral (t = pi/4 to 7pi/4 ) (r^2/2)(r=0 to 1)dt
= Integral (t = pi/4 to 7pi/4 ) (1/2)dt
= 1/2*(t)(t = pi/4 to 7pi/4)
= 1/2*[7pi/4 - pi/4]
= 3pi/4 cubic units ---------> This is the required volume
c> Surface area = area of the complete circle - area of the quarter wedge
= pi(1)^2 - 1/4*pi(1)^2
= pi - pi/4 = 3pi/4 square units
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