Department of Mathematics
Dr. Mustafa El-Agamy
Mathematics
for Engineering Students (Math 203)
WorkSheet No. (8) – Solution
Evaluate the following integrals:
1)
1
2
∫ ∫(𝑥 2 − 𝑦 2 ) 𝑑𝑦 𝑑𝑥
−1 −2
1
1
2
𝑦3
(𝟐)3
(−𝟐)3
2
2
2
= ∫ [𝑥 𝑦 − ] 𝑑𝑥 = ∫ [𝑥 (𝟐) −
] − [𝑥 (−𝟐) −
] 𝑑𝑥
3 −2
3
3
−1
−1
1
16
4 3 16 1
= ∫ 4𝑥 − 𝑑𝑥 = [ 𝑥 − 𝑥] = −𝟖
3
3
3 −1
2
−1
2)
1
√𝑦
∫ ∫ 𝑥 𝑦 𝑑𝑥 𝑑𝑦
0
0
1
1
𝑦
1
1
𝑥2 √
1
1
1 𝑦3
𝟏
2
2
2
= ∫ [𝑦 ] 𝑑𝑦 = ∫ [𝑦(√𝒚) ] − [ 𝑦(𝟎) ] 𝑑𝑦 = ∫ 𝑦 𝑑𝑦 = [ ] =
2 0
2
2
2 3 0 𝟔
0
3)
0
0
𝑥2
2
∫ ∫ (𝑥 + 2𝑦) 𝑑𝑦 𝑑𝑥
1
𝑥
2
= ∫ [𝑥𝑦 +
1
2
𝑦 2 ]𝑥𝑥
2
𝑑𝑥 = ∫ [𝑥(𝒙𝟐 ) + (𝒙𝟐 )2 ] − [𝑥(𝒙) + (𝒙)2 ] 𝑑𝑥
1
2
2
𝑥5 𝑥4 2 𝑥3
𝟑𝟏𝟕
4
3
2
= ∫ 𝑥 + 𝑥 − 2𝑥 𝑑𝑥 = [ + −
] =
5
4
3 1
𝟔𝟎
1
4)
∬ 𝑥 2 𝑦 3 𝑑𝐴 ,
𝐷: 𝑦 = 2𝑥, 𝑥 = 1 and 𝑦 = 0
𝐷
Draw the boundaries of the region 𝐷
𝑥=1
𝑦 = 2𝑥
𝑦=0
1
2𝑥
2 3
∬ 𝑥 𝑦 𝑑𝐴 = ∫ ∫ 𝑥 2 𝑦 3 𝑑𝑦 𝑑𝑥
𝐷
0
1
0
2𝑥
𝑥 2𝑦4
𝟏 1 2
=∫ [
] 𝑑𝑥 = ∫ [𝑥 (𝟐𝒙)4 ] − [𝑥 2 (𝟎)4 ] 𝑑𝑥
4 0
𝟒 0
0
1
1
𝑥7
𝟒
= 𝟒 ∫ 𝑥 𝑑𝑥 = 4 [ ] =
7 0 𝟕
0
6
5)
∬ 2𝑦𝑒 𝑥 𝑑𝐴 ,
𝐷: 𝑦 = 𝑒 𝑥 , 𝑦 = 𝑒 and 𝑥 = 0
𝐷
Draw the boundaries of the region 𝐷
𝑦 = 𝑒𝑥
𝑦=𝑒
𝑥=0
𝑒
ln 𝑦
𝑥
2𝑦𝑒 𝑥 𝑑𝑥𝑑𝑦
∬ 2𝑦𝑒 𝑑𝐴 = ∫ ∫
𝐷
1
0
𝑒
= 2∫
1
𝑦
[𝑦𝑒 𝑥 ]ln
0
𝑒
𝑑𝑦 = 2 ∫ (𝑦𝑒 𝐥𝐧 𝒚 ) − (𝑦𝑒 𝟎 ) 𝑑𝑦
1
𝑒
𝑒
𝑦3 𝑦2
𝒆3 𝒆2
𝟏3 𝟏2
= 2 ∫ 𝑦 − 𝑦 𝑑𝑦 = 2 [ − ] = 2 [( − ) − ( − )] = 𝟔. 𝟑𝟑𝟓
3
2 1
3
2
3
2
1
2
6)
D is bounded by: 𝑦 = 𝑥 2 , 𝑦 = 0 and 𝑥 = 1
∬ 𝑥 cos 𝑦 𝑑𝐴 ,
𝐷
1 𝑥2
∬ 𝑥 cos 𝑦 𝑑𝐴 = ∫ ∫ 𝑥 cos 𝑦 𝑑𝑦 𝑑𝑥
𝐷
1 𝑥2
0 0
1
1
1
𝑥2
∫ ∫ 𝑥 cos 𝑦 𝑑𝑦 𝑑𝑥 = ∫[𝑥 𝐬𝐢𝐧 𝒚]0 𝑑𝑥 = ∫(𝑥 sin 𝑥 2 ) − (0) 𝑑𝑥 = ∫ 𝑥 sin 𝑥 2 𝑑𝑥
0 0
0
0
1
=
0
𝟏
1
1
∫ 𝟐𝑥 sin 𝑥 2 𝑑𝑥 = [− cos 𝑥 2 ]10 = ((− cos 1) − (− cos 0)) = 𝟎. 𝟐𝟑
𝟐
2
2
0
7)
∬ 3 − 6𝑥𝑦 𝑑𝐴 where 𝐷 is the region shown below.
𝐷
∴ ∬ 3 − 6𝑥𝑦 𝑑𝐴 = ∬ 3 − 6𝑥𝑦 𝑑𝐴 + ∬ 3 − 6𝑥𝑦 𝑑𝐴
𝐷
𝐷1
𝐷2
𝐷1
𝐷2
1
2 −𝑥 2
1
∬ 3 − 6𝑥𝑦 𝑑𝐴 = ∫ ∫ 3 − 6𝑥𝑦 𝑑𝑦 𝑑𝑥 + ∫ ∫ 3 − 6𝑥𝑦 𝑑𝑦 𝑑𝑥
−1 𝑥 2
𝐷
−2 −4
Calculate first integral
1
1
∫ ∫ 3 − 6𝑥𝑦 𝑑𝑦 𝑑𝑥
−1 𝑥 2
1
1
= ∫[3𝑦 − 3𝑥𝑦 2 ]1𝑥 2 𝑑𝑥 = ∫[3(𝟏) − 3𝑥(𝟏)2 ] − [3(𝒙𝟐 ) − 3𝑥(𝒙𝟐 )2 ] 𝑑𝑥
−1
−1
1
1
𝑥6 𝑥3 𝑥2
5
2
= ∫ 3𝑥 − 3𝑥 − 3𝑥 + 3 𝑑𝑥 = 3 [ − − + 𝑥] = 𝟒
6
3
2
−1
−1
Calculate second integral
2 −𝑥 2
∫ ∫ 3 − 6𝑥𝑦 𝑑𝑦 𝑑𝑥
−2 −4
2
2
2
𝟐
𝟐 2
2
= ∫[3𝑦 − 3𝑥𝑦 2 ]−𝑥
−4 𝑑𝑥 = ∫[3(−𝒙 ) − 3𝑥(−𝒙 ) ] − [3(−𝟒) − 3𝑥(−𝟒) ] 𝑑𝑥
−2
−2
2
2
−𝑥 6
= ∫ −3𝑥 − 3𝑥 + 48𝑥 + 12𝑑𝑥 = [
− 𝑥 3 − 24𝑥 2 + 12𝑥] = 𝟑𝟐
2
−2
5
−2
2
∴ ∬ 3 − 6𝑥𝑦 𝑑𝐴 = 4 + 32 = 𝟑𝟔
𝐷
8)
2
∬ 𝑒 𝑥 𝑑𝐴 ,
𝐷: 𝑦 = 2𝑥, 𝑥 = 1 and 𝑦 = 0
𝐷
𝑥=1
𝑦 = 2𝑥
𝑦=0
∬𝑒
𝑥2
1
2𝑥
2
𝑑𝐴 = ∫ ∫ 𝑒 𝑥 𝑑𝑦 𝑑𝑥
0
𝐷
1
0
𝑥2
2𝑥
1
𝑥2
1
2
= ∫ [𝑦𝑒 ]0 𝑑𝑥 = ∫ 𝑒 (𝟐𝒙 − 𝟎) 𝑑𝑥 = ∫ 2𝑥𝑒 𝑥 𝑑𝑥
0
0
0
2
1
= [𝑒 𝑥 ]0 = 𝑒 − 1 = 𝟏. 𝟕𝟐
9)
∬ cos(𝑦 3 ) 𝑑𝐴 ,
𝐷: 𝑦 = √𝑥, 𝑦 = 1 and 𝑥 = 0
𝐷
𝑥=0
𝑦 = √𝑥
𝑦=1
1
𝑦2
∬ cos(𝑦 3 ) 𝑑𝐴 = ∫ ∫ cos(𝑦 3 ) 𝑑𝑥 𝑑𝑦
𝐷
0
0
1
= ∫ [𝑥 cos(𝑦
0
3
𝑦2
)]0
1
1
3
𝑑𝑦 = ∫ cos(𝑦 ) (𝒚 − 𝟎) 𝑑𝑦 = ∫ 𝑦 2 cos(𝑦 3 ) 𝑑𝑦
0
𝟐
0
𝟏 1 2
1
1
= ∫ 𝟑𝑦 cos(𝑦 3 ) 𝑑𝑦 = [sin(𝑦 3 )]10 = sin(1) = 𝟎. 𝟐𝟖
𝟑 0
3
3
10)
4
∬ 𝑒 𝑦 𝑑𝐴 where 𝐷 is the region shown below.
𝐷
𝐷1
𝐷2
4
4
4
∬ 𝑒 𝑦 𝑑𝐴 = ∬ 𝑒 𝑦 𝑑𝐴 + ∬ 𝑒 𝑦 𝑑𝐴
𝐷
𝐷1
𝐷2
3
1 𝑦
∬𝑒
𝑦4
0
𝑑𝐴 = ∫ ∫ 𝑒
𝐷
𝑦4
0
4
𝑑𝑥 𝑑𝑦 + ∫ ∫ 𝑒 𝑦 𝑑𝑥 𝑑𝑦
−1 𝑦 3
0 0
There is a symmetry of the region and function. So, we can calculate only one integral.
3
1 𝑦
∫∫ 𝑒
0 0
1
𝑦4
𝑦4
𝑦3
1
1
𝑦4
4
𝑑𝑥 𝑑𝑦 = ∫[𝑥𝑒 ]0 𝑑𝑦 = ∫ 𝑒 (𝒚𝟑 − 𝟎) 𝑑𝑦 = ∫ 𝑦 3 𝑒 𝑦 𝑑𝑦
0
0
0
1
𝟏
1 𝑦4 1 1
3 𝑦4
= ∫ 𝟒𝑦 𝑒 𝑑𝑦 = [𝑒 ] = (𝑒 − 1)
𝟒
4
4
0
0
1
1
4
∴ ∬ 𝑒 𝑦 𝑑𝐴 = 2 ∗ (𝑒 − 1) = (𝑒 − 1) = 𝟎. 𝟖𝟔
4
2
𝐷
11) a)
𝐷: 𝑥 2 + 𝑦 2 = 4
∬ √𝑥 2 + 𝑦 2 𝑑𝐴 ,
𝐷
The domain is a circle. So, it will be better to use the polar coordinates instead of the
Cartesian coordinates.
Transformation from Cartesian to polar coordinates:
• 𝑥 = 𝑟 cos 𝜃
• 𝑥2 + 𝑦2 = 𝑟2
• 𝑦 = 𝑟 sin 𝜃
• 𝑑𝐴 = 𝑟 𝑑𝑟 𝑑𝜃
∬ √𝑥 2 + 𝑦 2 𝑑𝐴 = ∬ √𝒓𝟐 𝒓 𝒅𝒓 𝒅𝜽 = ∬ 𝑟 2 𝑑𝑟 𝑑𝜃
𝐷
𝐷
𝐷
2𝜋
= ∫
0
2
2𝜋
2
∫ 𝑟 𝑑𝑟 𝑑𝜃 = ∫
0
0
=
𝟏𝟔
𝝅
𝟑
2
2𝜋
𝑟3
𝟖
8
𝑑𝜃 = (2𝜋)
[ ] 𝑑𝜃 = ∫
3 0
3
0 𝟑
b)
∬ √𝑥 2 + 𝑦 2 𝑑𝐴,
𝐷: 1 ≤ 𝑥 2 + 𝑦 2 ≤ 4 in 1st quad.
𝐷
∬ √𝑥 2
𝐷
+
𝑦2
𝜋
2
𝟐
2
𝑑𝐴 = ∬ 𝒓 𝒅𝒓 𝒅𝜽 = ∫ ∫ 𝑟 2 𝑑𝑟 𝑑𝜃
0
𝐷
𝜋
2
2
1
𝜋
27
𝒓𝟑
𝟕
= ∫ [ ] 𝑑𝜃 = = ∫ 𝑑𝜃 = 𝝅
𝟑 1
𝟔
0
0 3
c)
𝐷: 𝑥 2 + 𝑦 2 = 2𝑥
∬ √𝑥 2 + 𝑦 2 𝑑𝐴,
𝐷
𝑟 is not a constant.
We will transform the equation of the circle to the polar form:
𝑥 2 + 𝑦 2 = 2𝑥 → 𝑟 2 = 2 𝑟 cos 𝜃 → 𝒓 = 𝟐 𝐜𝐨𝐬 𝜽
∬ √𝑥 2 + 𝑦 2 𝑑𝐴 = ∬ 𝒓𝟐 𝒅𝒓𝒅𝜽
𝐷
𝐷
𝜋
2
2 cos 𝜃
= ∫ ∫
𝜋
−
0
2
𝜋
2 8
𝜋
2
2 cos 𝜃
𝒓𝟑
2
𝑟 𝑑𝑟𝑑𝜃 = ∫ [ ]
𝜋 𝟑
−
0
𝜋
2
2
𝜋
2
𝟐𝟑 𝐜𝐨𝐬𝟑 𝜽
𝑑𝜃 = ∫
𝑑𝜃
𝜋
3
−
2
8
cos 3 𝜃 𝑑𝜃 = ∫ cos 𝜃 cos 2 𝜃 𝑑𝜃
𝜋
3 −𝜋
− 3
= ∫
2
𝜋
2
2
𝜋
8
8 2
= ∫ cos 𝜃 (𝟏 − 𝐬𝐢𝐧𝟐 𝜽) 𝑑𝜃 = ∫ cos 𝜃 − cos 𝜃 sin2 𝜃 𝑑𝜃
3 −𝜋
3 −𝜋
2
2
𝜋
8
sin3 𝜃 2
𝟑𝟐
= [sin 𝜃 −
] =
3
3 −𝜋
𝟗
2
12) a)
∬ 𝑥 𝑑𝐴
𝐷
𝐷 is bounded by: 𝑦 = √25 − 𝑥 2 , 3𝑥 − 4𝑦 = 0, 𝑦 = 0 and 1𝑠𝑡 𝑞𝑢𝑎𝑑.
3 √25−𝑦
∬ 𝑥 𝑑𝐴 = ∫ ∫
𝑅
0
3 √25−𝑦
∫ ∫
0
4
𝑦
3
2
2
𝑥 𝑑𝑥 𝑑𝑦
4
𝑦
3
3
√25−𝑦 2
𝑥2
𝑥 𝑑𝑥 𝑑𝑦 = ∫ [ ]
2 4𝑦
0
3
3
3
1
𝟏𝟔
𝑑𝑦 = ∫(𝟐𝟓 − 𝒚𝟐 ) − ( 𝒚𝟐 ) 𝑑𝑦
2
𝟗
0
1
25 2
25
1 3 3
= ∫ 25 − 𝑦 𝑑𝑦 =
[𝑦 − 𝑦 ] = 𝟐𝟓
2
9
2
27
0
0
b)
∬ 𝑥 𝑑𝐴 ,
𝐷 is bounded by: 𝑥 2 + 𝑦 2 = 9 , 𝑦 = 𝑥 and 1st quad
𝐷
∬ 𝑥 𝑑𝐴 = ∬ 𝒓 𝐜𝐨𝐬 𝜽 𝒓 𝒅𝒓 𝒅𝜽 = ∬ 𝑟 2 cos 𝜃 𝑑𝑟 𝑑𝜃
𝐷
𝐷
𝜋
4
𝐷
𝜋
4
3
3
𝜋
4 (𝟑𝟑 − 𝟎𝟑 )
𝒓𝟑
2
∫ ∫ 𝑟 cos 𝜃 𝑑𝑟𝑑𝜃 = ∫ [ cos 𝜃] 𝑑𝜃 = ∫
cos 𝜃 𝑑𝜃
𝟑
3
0 0
0
0
0
𝜋
4
= ∫ 9 cos 𝜃 𝑑𝜃 = 9 [𝐬𝐢𝐧 𝜽]30 = 𝟗 𝐬𝐢𝐧 𝟑
0
c)
∬ 𝑥 𝑑𝐴 ,
𝐷 is bounded by 𝑥 2 + 𝑦 2 = 2𝑦
𝐷
∬ 𝑥 𝑑𝐴 = ∬ 𝒓 𝐜𝐨𝐬 𝜽 𝒓 𝒅𝒓𝒅𝜽 = ∬ 𝑟 2 cos 𝜃 𝑑𝑟𝑑𝜃
𝐷
𝐷
𝐷
𝑟 is not a constant
We will transform the equation of the circle to the polar form:
𝑥 2 + 𝑦 2 = 2𝑦 → 𝑟 2 = 2 𝑟 sin 𝜃 → 𝒓 = 𝟐 𝐬𝐢𝐧 𝜽
𝜋
2 sin 𝜃
2
∬ 𝑟 cos 𝜃 𝑑𝑟𝑑𝜃 = ∫ ∫
𝐷
0
𝜋
𝑟 2 cos 𝜃 𝑑𝑟𝑑𝜃
0
2 sin 𝜃
𝒓𝟑
= ∫ [ cos 𝜃]
𝟑
0
0
𝜋
= ∫
0
𝜋
𝟐𝟑 𝐬𝐢𝐧𝟑 𝜽
𝑑𝜃 = ∫
cos 𝜃 𝑑𝜃
3
0
8 3
8
[𝐬𝐢𝐧𝟒 𝜽]𝜋0 = 𝟎
sin 𝜃 cos 𝜃 𝑑𝜃 =
3
3∗𝟒
13)
1
1
∫ ∫ √𝑥 3 + 1 𝑑𝑥 𝑑𝑦
0
√𝑦
√𝑥 3 + 1 does not have antiderivative for 𝑥. We have to interchange order of integration.
𝑦=1
∫
𝑥=1
√𝑥 3 + 1 𝑑𝑥 𝑑𝑦
∫
𝑦=0
𝑥=√𝑦
• 𝑥=1
• 𝑥 = √𝑦
• 𝑦: 0
→
→
𝑦 = 𝑥2
𝑦=1
1
𝑦 = 𝑥2
𝑥=1
𝑦=0
∴ By choosing vertical strip to interchange the order of integration
1
1
∫ ∫
0
1
√𝑥 3
+ 1 𝑑𝑥 𝑑𝑦 = ∫ ∫
√𝑦
0
1
1
𝑥2
√𝑥 3
+ 1 𝑑𝑦 𝑑𝑥 =
0
∫ [(√𝑥 3
+ 1) 𝑦]
0
1
𝑥2
0
𝑑𝑥
1
= ∫ (√𝑥 3 + 1 ) 𝑥 2 𝑑𝑥 = ∫ 𝑥 2 (𝑥 3 + 1)2 𝑑𝑥
0
0
1
=
1
𝟏
1
∫ 𝟑 𝑥 2 (𝑥 3 + 1)2 𝑑𝑥 = [
𝟑
3
0
(𝑥 3
3 1
1)2
+
3/2
] = 𝟎. 𝟒𝟎𝟔
0
14)
3
6
∫ ∫ √𝑦 2 + 2 𝑑𝑦 𝑑𝑥
0
2𝑥
∵ √𝑦 2 + 2 does not have antiderivative for 𝒚
𝑥=3
𝑦=6
∫
∫
𝑥=0
√𝑦 2 + 2 𝑑𝑦 𝑑𝑥
𝑦=2𝑥
• 𝑦 = 2𝑥
• 𝑦=6
• 𝑥: 0 → 3
∴ By choosing horizontal strip to interchange the order of integration
3
6
∫ ∫
0
1
𝑦
2
6
√𝑦 2
+ 2 𝑑𝑦 𝑑𝑥 = ∫ ∫
2𝑥
0
0
6
=
6
√𝑦 2
+ 2 𝑑𝑥 𝑑𝑦 = ∫ [𝑥√𝑦 2 + 2]
0
1
1
1
∫ ( . 𝟐𝑦√𝑦 2 + 2) 𝑑𝑦 = [
𝟐 0 2
4
(𝑦 2
𝑦/2
𝑑𝑦
0
3 6
2)2
+
3/2
] = 𝟑𝟖. 𝟔
0
15)
1
1
2
∫ ∫ 𝑥 𝑒 𝑦 𝑑𝑦 𝑑𝑥
𝑥2
0
2
∵ 𝑥 𝑒 𝑦 does not have antiderivative for 𝒚.
𝑥=1
∫
𝑦=1
2
𝑥 𝑒 𝑦 𝑑𝑦 𝑑𝑥
∫
𝑦=𝑥 2
𝑥=0
• 𝑦=1
• 𝑦 = 𝑥2
• 𝑥: 0 → 1
∴ By choosing horizontal strip to interchange the order of integration
1
1
∫ ∫ 𝑥𝑒
0
𝑥2
𝑦2
1
√𝑦
2
𝑑𝑦 𝑑𝑥 = ∫ ∫ 𝑥 𝑒 𝑦 𝑑𝑥 𝑑𝑦
0
0
𝑦
𝑥 2 𝑦2 √
1 1
1 1 1
2
2
𝑦2
= ∫ [ 𝑒 ] 𝑑𝑦 = ∫ (√𝑦 ) 𝑒 𝑑𝑦 = . ∫ 𝟐𝑦 𝑒 𝑦 𝑑𝑦
2
2 0
2 𝟐 0
0
0
1
1
1
2 1
= [𝑒 𝑦 ] = (𝑒 − 1) = 𝟎. 𝟒𝟑
4
0 4
√8
16)
∫
2
5𝑥 3 cos(𝑦 3 ) 𝑑𝑦 𝑑𝑥
∫
𝑥 2 ⁄4
0
∵ 5𝑥 3 cos(𝑦 3 ) does not have antiderivative for 𝒚
𝑥=√8
𝑦=2
∫
5𝑥 3 cos(𝑦 3 ) 𝑑𝑦 𝑑𝑥
∫
𝑦=𝑥 2 /4
𝑥=0
• 𝑦 = 𝑥 2 /4
• 𝑦=2
• 𝑥: 0 → √8
∴ By choosing horizontal strip to interchange the order of integration
√8
∫
2
∫
2
3
5𝑥 cos(𝑦
3)
𝑑𝑦 𝑑𝑥 = ∫ ∫
𝑥 2 ⁄4
0
2√𝑦
0
5𝑥 3 cos(𝑦 3 ) 𝑑𝑥 𝑑𝑦
0
2√ 𝑦
5 4
5 2
20 2 2
4
3)
3)
= ∫ [ 𝑥 cos(𝑦 ]
𝑑𝑦 = ∫ (2√𝑦 ) cos(𝑦 𝑑𝑦 =
∫ 𝟑𝑦 cos(𝑦 3 ) 𝑑𝑦
4 0
𝟑 0
0
0 4
2
=
20
2 20
[sin(𝑦 3 )] =
sin(8) = 𝟔. 𝟔
3
0
3
17)
3
√9−𝑥 2
∫ ∫
𝑒 (𝑥
2 +𝑦 2 )
𝑑𝑦 𝑑𝑥
−3 0
• 𝑦=0
• 𝑦 = √9 − 𝑥 2 → 𝑥 2 + 𝑦 2 = 9
• 𝑥: − 3 → 3
3
√9−𝑥 2
∫ ∫
−3 0
𝑒
(𝑥 2 +𝑦 2 )
𝜋
3
1 𝜋 3 𝒓𝟐
𝑑𝑦 𝑑𝑥 = ∫ ∫ 𝒆 𝒓 𝒅𝒓 𝒅𝜽 = ∫ ∫ 𝒆 𝟐𝒓 𝒅𝒓 𝒅𝜽
𝟐 0 0
0 0
𝒓𝟐
1 𝜋 𝑟2 3
1 𝜋 9
𝜋
= ∫ [𝑒 ] 𝑑𝜃 = ∫ (𝑒 − 1) 𝑑𝜃 = (𝑒 9 − 1)
2 0
0
2 0
2
= 𝟏𝟐𝟕𝟐𝟔. 𝟕
18)
1
0
cos (√𝑥 2 + 𝑦 2 ) 𝑑𝑥 𝑑𝑦
∫ ∫
0
− √1−𝑦 2
• 𝑥=0
• 𝑥 = − √1 − 𝑦 2 → 𝑥 2 + 𝑦 2 = 1
• 𝑦: 0 → 1
1
0
∫ ∫
0
𝜋
cos (√𝑥 2
+
𝑦2
) 𝑑𝑥 𝑑𝑦 = ∫
− √1−𝑦 2
1
∫ 𝐜𝐨𝐬(𝒓) 𝒓 𝒅𝒓 𝒅𝜽
𝜋/2 0
𝑑
𝑑𝑟
𝜋
1
𝑟
cos(𝑟)
1
sin(𝑟)
0
− cos(𝑟)
𝜋
∫ ∫ cos(𝑟) 𝑟 𝑑𝑟 𝑑𝜃 = ∫ [𝑟 sin(𝑟) + cos(𝑟)]
𝜋
2
0
𝜋
2
∫
1
𝑑𝜃
0
𝜋
= ∫ sin(1) + cos(1) − 1 𝑑𝜃 =
𝜋
2
𝜋
[sin(1) + cos(1) − 1] = 𝟎. 𝟔
2
19)
√2
∫
𝑥
√4−𝑥 2
2
∫ (𝑥 2 + 𝑦 2 )3⁄2 𝑑𝑦 𝑑𝑥 + ∫ ∫
0
0
(𝑥 2 + 𝑦 2 )3⁄2 𝑑𝑦 𝑑𝑥
√2 0
∵ (𝑥 2 + 𝑦 2 )3⁄2 does not have antiderivative neither for 𝒙 nor 𝒚.
• 𝑦=𝑥
• 𝑦=0
• 𝑥: 0 → √2
• 𝑦 = √4 − 𝑥 2 → 𝑥 2 + 𝑦 2 = 4
• 𝑦=0
• 𝑥: √2 → 2
√2
∫
0
𝑥
2
√4−𝑥 2
∫ (𝑥 2 + 𝑦 2 )3⁄2 𝑑𝑦 𝑑𝑥 + ∫ ∫
0
(𝑥 2 + 𝑦 2 )3⁄2 𝑑𝑦 𝑑𝑥
√2 0
𝜋/4
2
∫ (𝒓𝟐 )𝟑/𝟐 𝒓 𝒅𝒓 𝒅𝜽
=∫
0
𝜋/4
=∫
0
0
2
𝜋/4
∫ 𝒓𝟒 𝒅𝒓 𝒅𝜽 = ∫
0
0
2
𝑟5
1 𝜋/4
𝟖
[ ] 𝑑𝜃 = ∫ 32 𝑑𝜃 = 𝝅
5 0
5 0
𝟓
20. Find the area between the curves 𝑦 = 𝑥 2 and 𝑦 = 2𝑥.
Solution:
Double integration used in calculating the area : 𝐴 = ∬ 1 𝑑𝐴
𝐷
2
2𝑥
2
𝐴 = ∬ 1 𝑑𝐴 = ∫ ∫ 1 𝑑𝑦 𝑑𝑥 = ∫ [ 𝒚
𝐷
0
𝑥2
0
]2𝑥
𝑥2
2
𝑑𝑥 = ∫ 2𝑥 − 𝑥 2 𝑑𝑥
0
2
𝒙𝟑
𝟒
𝟐
= [𝒙 − ] =
𝟑 0 𝟑
21. Use double integration to find the volume of the solid that is bounded above by the cone
𝑧 = 4 − √𝑥 2 + 𝑦 2 and below by the 𝑥𝑦 − plane.
Solution:
Double integration used in calculating the volume: 𝑉 = ∬(𝑧2 − 𝑧1 ) 𝑑𝐴
𝐷
∴ 𝑉 = ∬ (4 − √𝑥 2 + 𝑦 2 ) − (0) 𝑑𝐴
𝐷
⸫ The domain is the projection of the surface 𝑧 = 4 − √𝑥 2 + 𝑦 2 onto the 𝑥𝑦-plane
• 4 − √𝑥 2 + 𝑦 2 = 0 → 𝑥 2 + 𝑦 2 = 16
2𝜋
𝑉 = ∬4−
𝐷
2𝜋
𝑉= ∫
0
2𝜋
= ∫
0
√𝑥 2
+
𝑦2
𝑑𝐴 = ∫
0
4
∫ (4 − √𝒓𝟐 ) 𝒓 𝒅𝒓𝒅𝜽
0
4
∫ 4𝑟 − 𝑟 2 𝑑𝑟𝑑𝜃
0
4
2𝜋
2𝜋
(𝟒)𝟑
𝒓𝟑
32
𝟏𝟐𝟖
𝟐
𝑑𝜃 = ∫
𝑑𝜃 =
[𝟐𝒓 − ] 𝑑𝜃 = ∫ 𝟐(𝟒 ) −
𝟑 0
𝟑
3
𝟑
0
0
𝟐
22. The density at any point on the region 𝐷: 𝑥 2 + 𝑦 2 = 4 is given by 𝜌(𝑥, 𝑦) = √𝑥 2 + 𝑦 2 .
Use double integral to calculate the mass of that region.
Solution:
Double integration used in calculating the mass : 𝑀 = ∬ 𝜌(𝑥, 𝑦) 𝑑𝐴
𝐷
𝑀 = ∬ √𝑥 2 + 𝑦 2 𝑑𝐴 = ∬ √𝒓𝟐 𝒓 𝒅𝒓𝒅𝜽 = ∬ 𝑟 2 𝑑𝑟𝑑𝜃
𝐷
𝐷
2𝜋
=∫
0
2
2𝜋
2
∫ 𝑟 𝑑𝑟𝑑𝜃 = ∫
0
0
𝐷
2
2𝜋
𝒓𝟑
8
𝟏𝟔
𝑑𝜃 =
𝝅
[ ] 𝑑𝜃 = ∫
𝟑 0
𝟑
0 3