Practice Exam 4 Solution

# Do not write q evaluate the integral z 8 z r3 y 1 1 x

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Unformatted text preview: 0 10 − 2x 5 dx 11. Let Q be the region between z = (x2 + y 2 )3/2 and z = 1, and inside x2 + y 2 = 4. Sketch the region Q, and then x2 + y 2 ez dV as an integral in the best(for this example) 3-dimensional coordinate system. DO NOT write Q EVALUATE THE INTEGRAL. z 8 z = r3 y 1 1 x 2 Given the form of the solid region and the function, cylindrical coordiates is the best system to use to express this integral. Converting the top and bottom surfaces, we have z = r2 3 2 = r3 and z = 1 Notice that the intersection between these two surfaces is: (x2 + y 2 )3/2 = 1, or x2 + y 2 = 1, the circle of radius 1 about the origin. The region in the plane is all points between the circle of radius one and the circle of radius 2 both centered at the origin, and the integrand becomes rez . r3 2 2π r3 2 2π r2 ez dz dr dθ (rez ) r dz dr dθ = Thus the integral is: 0 1 0 1 1 1 12. Set up a triple integral in spherical coordinates for the volume V of the region between z = 3x2 + 3y 2 and the sphere x2 + y 2 + z 2 = 16. Be sure to include a sketch of the region, and DO NOT EVALUATE THE INTEGRAL. z 4 ρ=4 3 y x 3x2 + 3y 2 , we have z 2 = 3x2 + 3y 2 , or 4z 2 = 3x2 + 3y 2 + 3z 2 = 3ρ2 Translating z = 2 Then 4 (ρ cos φ) = 3ρ2 , or 4ρ2 cos2 φ = 3ρ2 . Thus we have 4 cos2 φ = 3, or cos2 φ = 3 . 4 Hence cos φ = ± √ 3 2. 2 Since we only want values between 0 and π , we take φ = π 6, [a 30◦ cone]. Notice that x2 + y + z 2 = 16 is the sphere of readius 4 about the origin, and has the form ρ = 4. 2π 4 π 6 ρ2 sin φ dρ dφ dθ Therefore, the integral has the form 0 0 0 3x2 + 3y 2 and z = 13. (a) Let Q be the region between z = 4 − x2 − y 2 . Sketch the region Q z 2 ρ=2 3 y x (b) Set up, but do not evaluate a triple integral in rectangular coordinates that gives the volume of Q. √ 22 √ 2 1− x 1 1 dV = 4 Using symmetry, 4− x − y 1 dz dy dx 3x2 +3y 2 0 0 Q √ (c) Set up, but do not evaluate a triple integral in cylindrical coordinates that gives the volume of Q. 1 π 2 Again using symmetry, 1 dV = 4 Q √ 0 0 √ 4− r 2 1 r dz dr dθ 3r (d) Set up, but do not evaluate a triple integral in spherical coordinates that gives the volume of Q. We need to ﬁgure out the angle of declination for the cone. Starting with z 2 = 3x2 + 3y 2 , we see that 4z 2 = 3x2 + 3y 2 + 3z 2 , or 4z 2 = 3ρ2 . 3 Therefore, since z = ρ cos φ, 4ρ2 cos2 φ = 3ρ2 . Hence 4 cos2 φ = 3, or cos2 φ = . 4 √ 3 π . The solution to this in the ﬁrst quadrant is φ = . Thus cos φ = ± 2 6 2π 2 π 6 ρ2 sin φ dρ dφ dθ 1 dV = So we have 0 0 Q 0 (e) Pick one of the triple integrals you found above and evaluate it in order to ﬁnd the volume of Q exactly. 2π ρ sin φ dρ dφ dθ = 0 π 6 = 0 8 3 0 8 8 sin φ dφ dθ = 3 3 π 6 2π 0 − cos φ 0 8 dθ = 3 0 0 0 0 0 13 ρ sin φ 3 2 dφ dθ 0 π 6 2π 0 − cos φ − √ 3 −1 2 2π π 6 2 We will use the spherical integral: 2π 2π 2 π 6 dθ 0 8 dθ = 3 √ 1− 3 2 2π θ 0 √ 8 3 = (2π ) 1 − 3 2 16π = 3 √ 3 1− 2...
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## This document was uploaded on 03/11/2014 for the course MATH 323 at Minnesota State University Moorhead .

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