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math103Exam3solutions

math103Exam3solutions - Math 103 Exam 3 Instructions You...

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Math 103 - Exam 3 Instructions: You are not permitted to use a calculator or any other materials on this exam. Show all your work on the test. You may use the back of the pages if needed. Simplify all answers. Give exact answers (i.e. fractions, radicals, etc.) unless instructed otherwise. Sign the Honor Pledge after you finish the exam. I have neither given nor received any unauthorized help on this test, and I have conducted myself within the guidelines of the university honor code. Also I will not discuss the contents of this exam with any other person until after the exams are handed back in class. Name and section: Pledge: Question: 1 2 3 4 5 6 7 Total Points: 7 8 7 8 7 7 7 51 Score:
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1. (7 points) Find the volume of the solid with uniform density that lies inside the sphere ρ = a and above the cone φ = π 4 . Solution: Since the density is uniform, then we just need to integrate the constant δ = c . Here it will be easier to use spherical coordinates to integrate over T which region indicated in the problem. V = ZZZ T c dV = c Z 2 π 0 Z π 4 0 Z a 0 ρ 2 sin φ dρ dφ dθ = c Z 2 π 0 Z π 4 0 1 3 ρ 3 sin φ | a 0 dφ dθ = c Z 2 π 0 Z π 4 0 1 3 a 3 sin φ dφ dθ = c Z 2 π 0 - 1 3 a 3 cos φ | π 4 0 = c Z 2 π 0 - 2 6 a 3 + 1 3 a 3 = ca 3 ( 2 - 2 3 ) π 2. (8 points) Using integration show that the surface area of a sphere of radius R is 4 πR 2 . Solution: Recall that that dS = | r ∂u × r ∂v | du dv , so we first need a parametrization of the sphere. Using symmetry we need only consider the top half of the sphere which
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