Question 1 a.

Consider 𝜙 = Ѳ + Ѱ, 𝑡ℎ𝑒𝑛 𝑟 = 𝑓(Ѳ)
In polar coordinates 𝑥 = 𝑟𝑐𝑜𝑠 Ѳ 𝑎𝑛𝑑 𝑦 = 𝑟𝑠𝑖𝑛Ѳ
𝑑𝑥
𝑑𝑥
= −𝑟𝑠𝑖𝑛 Ѳ + 𝑦 = 𝑟𝑐𝑜𝑠Ѳ
,
𝑑Ѳ
𝑑Ѳ

The geometry of the problem is evident that 2𝜋 − ѡ − Ѳ = 2𝜋 − 𝜙, thus ѡ = 𝜙 − Ѳ, therefore
the trig formula can been used
tan Ѱ = tan(𝜙 − Ѳ) =

𝑑𝑦
𝑑𝑥
= 𝑟𝑐𝑜𝑠Ѳ + 𝑠𝑖𝑛Ѳ ,
𝑑Ѳ
𝑑Ѳ
𝑡𝑎𝑛𝜙 − 𝑡𝑎𝑛Ѳ
.
1 + 𝑡𝑎𝑛𝜙𝑡𝑎𝑛Ѳ
Since tangent line to curve 𝑓(Ѳ) makes and angle 𝜙 with respect to x-axis, we have
𝑡𝑎𝑛𝜙 =
𝑑𝑦/𝑑Ѳ
𝑎𝑛𝑑 𝑡ℎ𝑢𝑠 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎,
𝑑𝑦/𝑑 Ѳ
𝑡𝑎𝑛Ѳ =
𝑦
𝑔𝑖𝑣𝑒𝑠,
𝑥
𝑑𝑦/𝑑𝑥 𝑦
𝑑𝑦
𝑑𝑥
−
𝑥
−𝑦
𝑑𝑥/𝑑Ѳ 𝑥
𝑑Ѳ
𝑑Ѳ
tan Ѱ =
=
𝑦 𝑑𝑦/𝑑𝑥
𝑑𝑦
𝑑𝑥
1+( )
𝑥
+𝑦
𝑥 𝑑𝑥/𝑑Ѳ
𝑑Ѳ
𝑑Ѳ
𝑑𝑟
𝑑𝑟
) − 𝑦(−𝑟𝑠𝑖𝑛Ѳ + 𝑐𝑜𝑠Ѳ )
𝑑Ѳ
𝑑Ѳ
=
𝑑𝑟
𝑑𝑟
𝑥 (−𝑟𝑠𝑖𝑛Ѳ + 𝑐𝑜𝑠Ѳ ) + 𝑦 (𝑟𝑐𝑜𝑠Ѳ + 𝑠𝑖𝑛Ѳ )
𝑑Ѳ
𝑑Ѳ
𝑥 (𝑟𝑐𝑜𝑠Ѳ + 𝑠𝑖𝑛Ѳ
By substituting, we have,
tan Ѱ =
𝑑𝑟
𝑑𝑟
𝑟𝑐𝑜𝑠Ѳ (𝑟𝑐𝑜𝑠Ѳ + 𝑠𝑖𝑛Ѳ ( )) − 𝑟𝑠𝑖𝑛Ѳ (−𝑟𝑠𝑖𝑛Ѳ + 𝑐𝑜𝑠Ѳ ( ))
𝑑Ѳ
𝑑Ѳ
𝑑𝑟
𝑑𝑟
𝑟𝑠𝑖𝑛Ѳ (−𝑟𝑠𝑖𝑛Ѳ + 𝑐𝑜𝑠Ѳ ( )) + 𝑟𝑐𝑜𝑠Ѳ (𝑟𝑐𝑜𝑠Ѳ + 𝑠𝑖𝑛Ѳ ( ))
𝑑Ѳ
𝑑Ѳ
𝑟 2 𝑐𝑜𝑠 2 Ѳ + 𝑟 2 𝑠𝑖𝑛2 Ѳ
𝑟
=
𝑑𝑟
𝑑𝑟⁄
𝑟𝑐𝑜𝑠 2
𝑟 + 𝑟 2 𝑠𝑖𝑛2 Ѳ
𝑑Ѳ
𝑑Ѳ
𝑡𝑎𝑛Ѱ =
𝑟
𝑑𝑟/𝑑 Ѳ
Ѱ = 𝑡𝑎𝑛−1 (
𝑟
)
𝑑𝑟
Ѳ
𝑑
Question 1 c

the equation of the larger circle in the cartesian coordinate equals, 𝑥 2 + 𝑦 2 = 1

the points are, (1,

the center of the circle would be (

the equation of the smaller circle would be (𝑥 −

𝑥 2 + 𝑦 2 − √2𝑥 = 0

𝑙𝑒𝑡 𝑥 = 𝑅𝑐𝑜𝑠Ѳ, 𝑎𝑛𝑑 𝑦 = 𝑅𝑠𝑖𝑛Ѳ
𝜋
)
√2
1 1
,
√2 √2
𝑎𝑛𝑑 𝑒𝑞𝑢𝑎𝑙𝑠 (
)
1
, 0 ) 𝑎𝑛𝑑
√2
𝑖𝑡𝑠 𝑟𝑎𝑑𝑖𝑢𝑠 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒
1 2
)
√2
𝑥 2 = √2𝑥 𝑅𝑐𝑜𝑠Ѳ
𝑅 = √2𝑥 𝑐𝑜𝑠Ѳ , That is the equation of the small circle
+ 𝑦2 =
1
√2
1
√2
𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒
Question 3
a)
I.
𝑦
𝑑𝐴
𝑥 2 +𝑦 2
∬𝑅
.R→ 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑜𝑑 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑏𝑦 𝑦 = 𝑥; 𝑦 = 2𝑥 ; 𝑥 = 1 ; 𝑥 = 2.
𝑦
𝑦
2
2𝑥
5𝑥 2
𝑦
∫ ∫ − 𝑥2 +𝑦2 𝑑𝐴 = ∫1 ∫𝑥
𝑦
𝑑𝑦𝑑𝑥
𝑥 2 +𝑦2
𝐿𝑒𝑡 𝑢 = 𝑥 2 + 𝑦 2
2𝑥
= ∫𝑥
=
1
2
𝑦
𝑥 2 +𝑦 2
𝑑𝑦 = ∫2𝑥 2 2 2𝑢 𝑑𝑢
5𝑥 2
2
5𝑥
∫2𝑥 2 2 ∗ 𝑢 𝑑𝑢 = 2 [ln(𝑢)] 2
2𝑥
1
1
1
= 2 ln[5𝑥 2 ] − ln[2𝑥 2 ]
1
= 2 ln[5] − ln[2]
21
= ∫1 2 ln(5) − ln(2) 𝑑𝑥
21
= ∫1 2 ln(5) − ln(2) 𝑑𝑥
1
2
= 2 [ln(5) − ln(2) x]
1
1
5
= 2 (ln(2)(2 − 1)
𝟏
𝟓
Thus; R = 𝟐 𝐥𝐧(𝟐)
II.
∫ ∫𝑅 𝑥𝑑𝐴. Where R is a sector of the circle bounded by 𝑦 = √25 − 𝑥 2 , 3𝑥 − 4𝑦 = 0, 𝑦 = 0
3𝑥 − 4𝑦 = 0
3𝑥
4
𝑦=
𝑎𝑛𝑑 𝑥 =
4𝑦
3
2
𝑦 = √25 − 𝑥 2 ; 𝑦 = 25 − 𝑥 2
𝑥 = √25 − 𝑦 2
𝑥 = √25 − 𝑦 2 𝑎𝑛𝑑 𝑥 =
=
4𝑦
3
𝑥=
< 𝑥 < √25 − 𝑦 2
4𝑦
3
=
16𝑦 2
9
+ 𝑦 2 = 25
25𝑦 2
9
= 25;
25𝑦 2
25
𝑦2 =
225
;𝑦
25
= √9
𝑦=3
= 225
4𝑦
3
√25−𝑦 2
3
𝑁𝑜𝑤; ∫0 ∫4𝑦
𝑥𝑑𝑥𝑑𝑦
3
√25−𝑦2
= ∫4𝑦
3
1
√25 − 𝑦 2
]
𝑑𝑦
4𝑦
2
3 𝑥2
𝑥𝑑𝑥 = ∫0
3
3
= 2 ∫0 25 − 𝑦 2 −
1
= 2 ∫ 25 (1 −
16𝑦 2
𝑑𝑦
9
𝑦2
) 𝑑𝑦
9
𝑦3 3
− 27]
0
=
25
[𝑦
2
=
(3)3
25
(3 − 27 )
2
=
25
(3 −
2
=
25
(2)
2
1)
= 𝟐𝟓
III.
Inside the cardioid 𝑟 = 1 + 𝑐𝑜𝑠𝜃 and outside the circle 𝑟 = 3𝑐𝑜𝑠𝜃
𝜃
𝑟 = 1 + 𝑐𝑜𝑠𝜃
0
45
90
135
180
225
270
315
360
2
1.71
1
0.29
0
0.29
1
1.71
2
𝑟1 = 𝑟2
1 + 𝑐𝑜𝑠𝜃 = 3𝑐𝑜𝑠𝜃
1 + 𝑐𝑜𝑠𝜃 − 3𝑐𝑜𝑠𝜃 = 0
1
𝑐𝑜𝑠𝜃 = 2 =
𝜋
3
There are two area on the graph
𝐴𝑟𝑒𝑎 = 𝐴1 + 𝐴2
𝜋
1+𝑐𝑜𝑠𝜃
𝐴 = 2(∫03 𝑑𝜃 ∫0
𝜋
3𝑐𝑜𝑠𝜃
𝑟𝑑𝑟 + ∫𝜋2 𝑑𝜃 + ∫0
3
3𝑐𝑜𝑠𝜃
3
2.12
0
-2.12
-3
-2.12
0
2.12
3
𝑟𝑑𝑟)
𝜋
1+𝑐𝑜𝑠𝜃
𝐴1 = ∫04 ∫0
𝑟𝑑𝑟𝑑𝜃