Tutorial06-solutions

# In this case the above inequalities imply that the

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Unformatted text preview: θ sin ρ cos θ cos −ρ sin θ sin sin θ sin ρ sin θ cos ρ cos θ sin cos sin ρ 1 −2 0 2 ρ cos Expanding along the last row, we obtain ρ2 cos ρ 2 Thus the answer is π 2 π 2 1 4 0 0 0 sin cos + √ ρρ sin √ cos ρ ρ sin 2 ρ cos θ=4 · 12 2 sin √ ρρ =√ sin 2 cos π √ π 2 8π 2 · [−2 cos ]0 · = 2 5 5 :) 5 E Extra examples Find the volume of the wedge cut from the cylinder = − and = 0 for ≥ 0. E S + 2 = 1 by planes Applying polar coordinates, ﬁnd the area given by 2 Here 2 + 22 ≤2 2 2 − 2 2 + 2 ≥ 2 > 0. In polar coordinates, the region can be rewritten as 4 ≤2 2 ( 2 cos2 θ − 2 sin2 θ ); cos 2θ ≥ 2 ≥ 2 which is equivalent to 2 ≤2 2 Thus, we must have 2 2 cos 2θ ≥ 2 Or equivalently, cos 2θ ≥ 1 2 which implies that − π π + 2 π ≤ 2θ ≤ + 2 π 3 3 for any ∈ Z. We can take θ ∈ [−π π ] so that 2θ ∈ [−2π 2π ]). In this case, the above inequalities imply that the only possible ’s we can take here are = 0 ±1 and we have − π π ≤ 2θ ≤ 3 3 or...
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## This document was uploaded on 02/10/2014.

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