Polar coordinates 5C. 1 a. a 4. π = 0 (0) is a circle centre, 0. and radius. The area of the semicircle is π =. π a


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1 Polr coordintes 5C r cos Are cos d (cos + ) sin + () + 8 cos cos r cos is circle centre, nd rdius. The re of the semicircle is. 8 Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
2 b r (+ sin ) Are (+ sin + sin )d + sin + cos d sin + cos d cos sin Use cos sin. Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
3 c r sin Are sin d (cos )d sin sin sin + + (+ ) 8 Use cos sin. Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
4 d r cos Are cosd sin sin () e r tn Are tn d lnsec ln () ln ln or Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
5 f r Are d () g r (+ cos ) ( cos cos )d + + [ ] + sin + sin + + () Are (9+ cos + cos )d ( + ) Use cos cos. Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free. 5
6 Are ( cos cos ) d p + pq + q q q cos cos d p + pq + + p + q q pq cos cos d + + p + q q sin sin + pq + p + q ( p + q ) + + () Use cos cos. Are cos d cos d ( cos )d + sin + () + Use cos cos. Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
7 In order to find the re enclosed by single loop, we clculte d ( 5sin ) d r + + ( sin + ) 5sin d 5 + sin ( cos ) d cos sin + ( + 5 ) nd >, we solve to obtin 9 In order to hve ( ) 5 First we find the intersection points of these two curves. sin sin sin sin + kor + k kor (+ k) k or (+ k) ( k Z). Since we re working in the rnge, we only cre bout the intersections occurring t, nd. In fct we only hve positive vlue for sin between nd so the loop is only defined in this rnge. Note tht sin when, which mens tht it intersects the origin. Now we clculte the re to be A ( sin ) d ( sin) d + + ( cos ) d ( cos8 ) sin sin d Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free. 7
8 The points of intersection re given by + sin sin So sin which mens 5 rcsin ( ) nd. So the polr coordintes of intersection re (, ) nd (, 5 ). Since there is symmetry bout the verticl xis, we my compute the re s Are ( sin) d+ ( + sin) d ( 9sin ) d + ( sin + ) + sin d 9 ( cos ) d + + sin ( cos ) d + 9 9sin sin + cos centred t (, ) with rdius 5. 7 The set of points A z: rgz { z: z + i 5} define segment of circle, Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free. 8
9 7 b The eqution for circle centred t (, ) with rdius 5, in Crtesin coordintes is ( x ) + ( y+ ) 5. To find the re of the region bounded bya, nd, we convert into polr coordintes, obtining ( rcos ) + ( rsin + ) 5. This simplifies to r 8cos sin when r. Now we clculte A (8cos sin ) d ( cos 9cos sin + sin )d ((+ cos ) 8sin + 8( cos ))d ( + cos 8( [ ( sin ) + + sin )) ] centred t (,5) with rdius. 8 The set of points A z: rgz { z: z+ 5 i } define region of circle, Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free. 9
10 8 b To find the re of the region bounded bya, r cos + sin to clculte ( cos sin ) A + d (57cos sin ) 8cos + sin d (88 (+ cos ) sin + 5 (cos ) ) d ( cos 5 ( ) 85. [ ( sin ) + ( sin )) ] nd, we use the polr form Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
11 9 In order to find the re of the shded region, we must find the re of the sector bounded by the curve nd the lineoa, then subtrct the re of the tringle OAB. The vlue of t the point A cn + be found by solving r + cos in order to get. We now find the re of the sector bounded by the curve nd the line OA. Asector ( + cos ) d ( + cos + cos ) d ( + cos + (cos + ) ) d [( sin ( sin )) ] Now we find the re of the tringle OAB by using the formul AreOAB bsinc, where is the length of OA, b is the length of OB nd C is the ngle between O A nd OB. AreOAB bsinc + + sin.77. Thus, the re of the shded region is found to be A Are Are sector (sf) OAB Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
12 Note tht there is symmetry bout the verticl xis nd so we my compute one side nd multiply by. In order to find the re of the shded region, we first must find the re of the sector of r + sin between nd the right hnd side intersection point of the two curves. This intersection point occurs when + sin sin i.e. when. So now we find the re of the sector of r + sin for. We will denote this re A. A ( + sin) d ( + sin + sin ) d ( + sin + (cos ) ) d [( cos + ( sin )) ] This integrl hs included smll extr region we do not wnt (the red region in the imge) We find the vlue of the unwnted regions re (which we will denote A ) by 9 A ( sin ) d ( 9sin ) 9 d (cos )d [( sin ) ] 9.8. So the right hnd side of the shded region hs re A right.97. Remember tht this is only hlf of the totl region we wnt to find, so tht mens we just need to double A right in order to find A totl.79 ( d.p.) Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
13 Chllenge The fr left hnd side of the shell is point which occurs when nd so we cn set r k. left Similrly for the fr right hnd side of the shell, which occurs t, giving r k. Thus the horizontl dimeter is d 7 k cm, so we conclude tht k 7 b In order to find the totl re of the cross section, we must ensure tht ll ngles re covered exctly once. So we choose to integrte between. A d d ( 8) right Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.
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