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Tutorial06-solutions

# Tutorial06-solutions - Problem Set 6 MH2100 Solutions For...

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Problem Set 6 MH2100 Solutions For more practice : Stewart: Section 15.1; Section 15.2; Section 15.3, exercises 7-32, 37-54; Section 15.4, exercises 5-27, 29-41; Section 15.6, exercises 1-14, 21-24; Section 15.8, exercises 7-12, 17-30; Section 15.9, exercises 1-34. 1 Questions on understanding the lecture E±²³´µ¶² · Assume that a region D ⊂ R 2 is bounded by half-lines θ = α and θ = β and the graph of a non-negative function ± = ± ( θ ) in polar coordinates. Show that Area of D = 1 2 ¸ β α [ ± ( θ )] 2 ²θ³ S¹º»¼µ¹½ The area is ¸¸ D ²´²µ = ¸ β α ¸ ± ( θ ) 0 ±²±²θ = 1 2 ¸ β α ± 2 ( θ ) :) 1

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2 Calculating double and triple integrals E±²³´µ¶² · For the following double integrals, draw the region of integration, change the order of integration, and evaluate them: (a) ¸ 2 0 ¸ 2 ± 2 ² ³ 2 ´³´± S¹º»¼µ¹½ 0.5 1 1.5 2 0.5 1 1.5 2 ¸ 2 0 ¸ 2 ± 2 ² ³ 2 = ¸ 2 0 ¸ ³ 0 2 ² ³ 2 ´±´³ = ¸ 2 0 2 ³² ³ 2 ´³ = ² ³ 2 ¾ ¾ ¾ 2 0 = ² 4 - 1 µ :) (b) ¸ 1 0 ¸ 1 ³ 3 ±² ± 3 S¹º»¼µ¹½ 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 2
± 1 0 ± 1 ± 3 ²³ ² 3 ´²´± = ± 1 0 ± ² 0 3 ² 3 ´±´² = ± 1 0 3 ² 2 ³ ² 3 ´² = ³ ² 3 ² ² ² 1 0 = ³ - 1 µ :) (c) ± 1 0 ± 1 ² 3 4 + ± 3 S³´µ¶·³¸ 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 ± 1 0 ± 1 ² 3 4 + ± 3 = ± 1 0 ± ± 2 0 3 4 + ± 3 = ± 1 0 3 ± 2 4 + ± 3 ´± = ln(4 + ± 3 ) ² ² ² 1 0 = ln 5 4 µ :) E¹º»¼·½º ¾ Compute the volume of the solid bounded by the surfaces ² = ± 2 , ² = 4, ² + = 6, and ² + = 8. S³´µ¶·³¸ The solid is shown in Figure 1. Its volume is ¿ 4 0 ¿ ² - ² ¿ 8 6 ´¶´±´² = 64 3 . :) E¹º»¼·½º À For the triple integral ± 2 0 Á ± 4 2 0 Á ± 2 ± 2 + 2 ´² Â ´± Â ´¶· sketch the solid whose volume is taken and rewrite the integral as ± * * Ã± * * Ã± * * ´¶ Ä ´² Ä ´± . 3

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Figure 1: Region of integration in Exercise 3 Figure 2: Region of integration in Exercise 4 4
Figure 3: Region of integration in Exercise 5 S±²³´µ±¶ The solid is a quarter of the cone (see Figure 2), the integral would be · 2 0 · 2 ± · ² 2 2 0 ³´³²³±µ :) E¸¹º»µ¼¹ ½ Evaluate the triple integral ··· V ²± 2 ´ 3 ³²³±³´ , where the volume V is bounded by the surfaces ´ = , ± = ² , ² = 1, ´ = 0. S±²³´µ±¶ It is ¾ 1 0 ¾ ² 0 ¾ ²± 0 2 ´ 3 ³´³±³² = 1 364 . :) 3 Calculation with change of variables E¸¹º»µ¼¹ ¿ Evaluate the area enclosed by the curve ( ² 2 + ± 2 ) 2 = ( ² 3 - 3 2 ), where ¶ > 0 is a parameter. Hint : You can re-write the equations deﬁning the curve using polar coordinates, and then apply the result of Exercise 2. Next, you will have to take some care in ﬁnding the integration limits α and β . 5

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S±²³´µ±¶ In polar coordinates, the curve can be rewritten as ± 4 = ² ( ± 3 cos 3 θ - 3( ± cos θ )( ± sin θ ) 2 ) ³ Or, ± = ² (cos 3 θ - 3 cos θ sin 2 θ ) ³ Note that, we have cos(3 θ ) = cos( θ + 2 θ ) = cos θ cos(2 θ ) - sin θ sin(2 θ ) = cos θ (cos 2 θ - sin 2 θ ) - 2 sin 2 θ cos θ = cos 3 θ - 3 cos θ sin 2 θ³ Thus, the curve is given by ± = ² cos 3 θ in polar coordinates. Further, since ± ≥ 0, we must have cos 3 θ ≥ 0, that is, - π 2 + 2 π´ ≤ 3 θ ≤ π 2 + 2 π´ or - π 6 + 2 3 ≤ θ ≤ π 6
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