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Unformatted text preview: Solutions to Assignment 2 by Sacha Nandlall T.A. for MATH 264, McGill University Email: [email protected] Website: http://www.resanova.com/teaching/calculus/ Fall 2006, Version 2 Question 1 Question: Find the volume in the upper half space z ≥ 0 below the cone x 2 + y 2 = z 2 inside the cylinder x 2 + y 2 = 2 x . Solution: Denote the region described above as R . We want to find its volume, which is ZZZ R dV . We need simply apply the six-step method described in the Tutorial 1 slides to evaluate the integral. 1. Sketch the region The sketch is similar to the one for Question 2, with the x and y axes interchanged. 2. Choose a coordinate system For the same reasons given in Question 2, we’ll choose cylindrical. 3. Parametrize the region of integration R This means, find bounds on r , θ , and z . From the sketch, we can see that z goes from the xy-plane ( z = 0) to the cone. The cone in cylindrical is given by x 2 + y 2 = z 2 ⇒ r 2 = z 2 ⇒ z = ± r ⇒ z = r where the last implication follows since z ≥ 0. Thus 0 ≤ z ≤ r . 1 To obtain the bounds on r and θ , project the region onto the xy- plane (this is the standard trick for obtaining bounds on r and θ in cylindrical). From the sketch, the projection will simply be the disc bounded by the circle corresponding to the wall of the cylinder, which is drawn in Figure 1). The equation for this circle is the same as that of the cylinder: x 2 + y 2 = 2 x ⇒ r 2 = 2 r cos θ ⇒ r = 2 cos θ Hence 0 ≤ r ≤ 2 cos θ . Moreover, the drawing in Figure 1 shows that- π/ 2 ≤ θ ≤ π/ 2. Therefore, the complete parametrization for the region R is: R : ≤ r ≤ 2 cos θ- π/ 2 ≤ θ ≤ π/ 2 ≤ z ≤ r 4. Write out the integrand The integrand is simply 1....
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This note was uploaded on 12/21/2010 for the course MATH 262 taught by Professor Faber during the Spring '08 term at McGill.
- Spring '08