![ENGG MATHS-II
Put x2 + y2 = r2 , dydx = rdrdπ
Limits:
r : 0β β , ΞΈ: 0 β
Ο
2
β
eβr2
rdrdΞΈ
Ο
β«
β
β«
β
eβ(X2+y2)dydx = β«2 β«
0 0 0 0
Substitution: Put r2 = t , if r = 0 β t = 0, r = β β t = β
2rdr = dt t: 0β β
rdr =
ππ‘
Ο
β«2 β«
0 0 2
β
eβt dt
dΞΈ
Ο
2
β
eβr2
rdrdΞΈ = β«2 β«
0 0
1
2 β1 0
eβt β
Ο
0
= β«2 [ ] dΞΈ
= ββ 0
+ e )dΞΈ
Ο
β«2(βe
0
=
1
2
1
2
Ο
β«2(0 + 1)dΞΈ
0
(β΅ eββ = 0, e0 = 1)
1
2
Ο
= β«2 dΞΈ
0
2 0
Ο
=
1
(ΞΈ)2
=
1
(
Ο
β 0)
=
2 2
Ο
4
Example:
π±
ππ²ππ±
π² π±π+π²π
Change into polar co-ordinates and then evaluate β«
π
β«
π
π
Solution:
Given order ππ₯ππ¦ is in correct form.
DOUBLE INTEGRATION](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Double Integration examples of double integration with substitution.pptx
- 1. Change into polar co-ordinates and then evaluate β« β β« β πβ(π±π+π²π)ππ²ππ± π π Solution: Given order ππ¦ππ₯ is in correct form. Given limits are π¦ βΆ 0 β β , π₯ βΆ 0 β β Equations are π¦ = 0 , π¦ = β , π₯ = 0 , π₯ = β Replacement: ENGG MATHS-II 5.4 Change of variables in Double and Triple integrals: Evaluation of double integrals by changing Cartesian to polar co- ordinates: Working rule: Step:1 Check the given order whether it is correct or not. Step:2 Write the equations by using given limits. Step:3 By using the equations sketch the region of integration. Step:4 Replacement: put x = rcosπ , y = rsinπ , x2 + y2 = r2 and dxdy = rdrdπ Step:5 Find r limits(draw radial strip inside the region) and ΞΈ limits and evaluate the integral. Example: DOUBLE INTEGRATION
- 2. ENGG MATHS-II Put x2 + y2 = r2 , dydx = rdrdπ Limits: r : 0β β , ΞΈ: 0 β Ο 2 β eβr2 rdrdΞΈ Ο β« β β« β eβ(X2+y2)dydx = β«2 β« 0 0 0 0 Substitution: Put r2 = t , if r = 0 β t = 0, r = β β t = β 2rdr = dt t: 0β β rdr = ππ‘ Ο β«2 β« 0 0 2 β eβt dt dΞΈ Ο 2 β eβr2 rdrdΞΈ = β«2 β« 0 0 1 2 β1 0 eβt β Ο 0 = β«2 [ ] dΞΈ = ββ 0 + e )dΞΈ Ο β«2(βe 0 = 1 2 1 2 Ο β«2(0 + 1)dΞΈ 0 (β΅ eββ = 0, e0 = 1) 1 2 Ο = β«2 dΞΈ 0 2 0 Ο = 1 (ΞΈ)2 = 1 ( Ο β 0) = 2 2 Ο 4 Example: π± ππ²ππ± π² π±π+π²π Change into polar co-ordinates and then evaluate β« π β« π π Solution: Given order ππ₯ππ¦ is in correct form. DOUBLE INTEGRATION
- 3. ENGG MATHS-II Given limits are x : yβ π , y : 0β π Equations are x =y , x = π , y = 0 , y = π Replacement: Put x = rcosπ , π₯2 + π¦2 = π2 , ππ₯ππ¦ = πππππ Limits: r : 0 β π , π: 0 β π πππ π 4 β« β« π π π₯ π₯ +π¦ 2 2 ππ¦ππ₯ = β« β« 0 π¦ π2 πππππ π ππππ π π 4 ππ π π 0 0 0 π π = β«4[ππππ π]πππ π ππ π πππ π 0 π = β«4 ( 0 πππ π β 0) ππ π = π β«4 ππ 0 0 π = π(π)4 4 = π( π β 0) = ππ 4 Example: π±π π² βπ±π+π²π Evaluate β« π β« π π ππ±ππ² by changing into polar co-ordinates. Solution: Given order ππ₯ππ¦ is in correct form. Given limits are x : yβ π , y : 0β π DOUBLE INTEGRATION
- 4. ENGG MATHS-II Equations are x =y , x = π , y = 0 , y = π Replacement: Put x2 = r2cos2π , π₯2 + π¦2 = π2 => π = βπ₯2 + π¦2 , dxdy = rdrdπ Limits: r : 0 β π , π: 0 β π πππ π 4 β« β« π₯2 βπ₯ +π¦ 2 2 ππ₯ππ¦ = β« β« π π 0 π¦ π πππππ π π2πππ 2π π 4 π π π π 0 0 3 3 π = β«4 [π π πππ 2π] πππ π ππ 0 π3 3 3πππ π 2 0 π = β«4 ( 0 πππ π β 0) ππ = π3 1 3 2 π β«4 3 0 πππ π πππ π ππ = π3 1 π 3 0 πππ π β«4 ππ = π 0 β«4 π πππ ππ = π3 3 π3 3 0 π (πππ(π πππ + π‘πππ))4 = π3 [πππ (π ππ π + π‘ππ π ) β πππ(π ππ0 + π‘ππ0)] 3 4 4 3 = π3 [πππ(β2 + 1) β πππ(1 β 0)] 3 = π3 πππ(β2 + 1) Note: 1. x2 + y2 = r2cos2ΞΈ + r2sin2ΞΈ = r2 2. β« cos 2 0 ΞΈdΞΈ = β« 2 2 Ο Ο 2 0 sin ΞΈdΞΈ = Γ 1 Ο 2 2 3. β« 4 cos ΞΈdΞΈ = β« Ο 2 0 4 Ο 2 0 sin ΞΈdΞΈ = Γ Γ 3 1 Ο 4 2 2 4. β« 2 2 Ο 2 0 cos ΞΈsin ΞΈdΞΈ = Γ Γ 1 1 Ο 4 2 2 DOUBLE INTEGRATION
- 5. Example: By changing into polar co-ordinates and evaluate β« ππ β« βπππ±βπ±π (π±π + π²π)ππ²ππ± π π Solution: Given order ππ¦ππ₯ is in correct form. Given limits are y : 0β β2ππ₯ β π₯2 , x : 0β 2π Equations are y = 0 , y = β2ππ₯ β π₯2 , x = 0 , x = 2π y2 = 2ax β x2 x2 + y2 β 2ax = 0 is a circle with centre (a,0) and radius βaβ. Replacement: Put x2 + y2 = r2 , ππ₯ππ¦ = πππππ Limits: r : 0 β 2ππππ π , π: 0 β π 2 2ππππ π π2 Γ πππππ π β« 2π β« β2ππ₯βπ₯2 (π₯2 + π¦2)ππ¦ππ₯ = β«2 β« 0 0 0 0 3 π ππππ 2ππππ π π = β«2 β« 0 0 4 ] 0 r4 2acosΞΈ Ο = β«2 [ 0 dΞΈ 4 Ο 24a4cos4ΞΈ = β«2( 0 β 0) dΞΈ Ο 0 = 4a4 β«2 cos4ΞΈdΞΈ ENGG MATHS-II DOUBLE INTEGRATION
- 6. = 4a4 Γ 3 Γ 1 Γ Ο 4 2 2 (β΅ β« 4 Ο 2 0 cos ΞΈdΞΈ = Γ Γ 3 1 Ο 4 2 2 ) = 3Οa4 4 Example: π± π±π+π²π ππ±ππ² By changing into polar co-ordinates and evaluateβ« π β« βππ±βπ±π π π Solution: Given order ππ₯ππ¦ is in incorrect form. X X2+y2 dydx The correct form is ππ¦ππ₯ β β« 2 β« β2XβX2 0 0 Given limits are y : 0β β2π₯ β π₯2 , x : 0β 2 Equations are y = 0 , y = β2π₯ β π₯2 , x = 0 , x = 2 y2 = 2x β x2 x2 + y2 β 2x = 0 is a circle with centre (1,0) and radius β1β. Replacement: Put x = rcosπ, π₯2 + π¦2 = π2 , dxdy = rdrdπ Limits: r : 0 β 2πππ π , π: 0 β π 2 π₯ +π¦ 2 2 ππ¦ππ₯ = β« β« 2 β2π₯βπ₯2 π₯ β«0β«0 π2 2πππ π ππππ π Γ πππππ π 2 0 0 0 π 0 =β«2[ππππ π]2πππ π ππ π 0 = β«2(2πππ 2π β 0)ππ ENGG MATHS-II DOUBLE INTEGRATION
- 7. π 0 = 2 β«2 πππ 2π ππ = 2 Γ 1 Γ π 2 2 (β΅ β« 2 π 2 0 1 π 2 2 πππ πππ = Γ ) = π 2 Example: By changing into polar co-ordinates and evaluateβ« π β« βππβπ±π βπ±π + π²πππ²ππ± π π Solution: Given order ππ₯ππ¦ is in correct form. Given limits are y : 0β βπ2 β π₯2 , x : 0β π Equations are y = 0 , y = βπ2 β π₯2 , x = 0 , x = π y2 = a2 β x2 x2 + y2 = a2 is a circle with centre (0,0) and radius βaβ. Replacement: Put x2 + y2 = r2 β r = βx2 + y2 , ππ¦ππ₯ = πππππ Limits: r : 0 β π , π: 0 β π a r Γ rdrdΞΈ 2 Ο β« a β« βa2βX2 βx2 + y2dydx = β«2 β« 0 0 0 0 a r2drdΞΈ Ο = β«2 β« 0 0 3 Ο r3 a = β«2 [ 0 ] dΞΈ 0 ENGG MATHS-II DOUBLE INTEGRATION
- 8. a3 3 Ο = β«2 ( 0 β 0)dΞΈ = a3 3 Ο β«2 dΞΈ 0 a3 3 0 Ο = (ΞΈ)2 3 2 = a3 ( Ο β 0) = Οa3 6 Example: π±π+π²π Evaluate β¬ π±ππ²π ππ±ππ² over the annular region between the circles x2 + y2 = a2 and x2 + y2 = b2 (b>a) by transforming into polar co-ordinates. Solution: π₯2+π¦2 π2 Replacement: Put x2 = r2cos2ΞΈ,y2 = r2sin2ΞΈ x2 + y2 = r2, ππ₯ππ¦ = πππππ Given the region is between the circles x2 + y2 = a2 and x2 + y2 = b2 Limits: r : π β π , π: 0 β 2π β΄ β¬ π₯2π¦2 ππ₯ππ¦ = β« 2π β« π π2πππ 2πΓπ2π ππ2π Γ πππππ 0 π ENGG MATHS-II DOUBLE INTEGRATION
- 9. r2 0 a ENGG MATHS-II = β« 2Ο β« b r5cos2ΞΈΓsin2ΞΈ Γ drdΞΈ = β« 2Ο β« b r3cos2ΞΈ Γ sin2ΞΈ drdΞΈ 0 a 4 a r4 b 2Ο 2 2 = β«0 [ ] cos ΞΈ Γ sin ΞΈ dΞΈ 4 0 = 1 β« 2Ο (b4 β a4) cos2ΞΈ Γ sin2ΞΈ dΞΈ 4 = (b4βa4) β« 2Ο cos2ΞΈ Γ sin2ΞΈ dΞΈ 0 = 4 4 (b βa ) 4 Γ β« Ο 2 4 0 2Ο 0 cos2ΞΈ Γ sin2ΞΈ dΞΈ (β΅ β« = 4β« ) Ο 2 0 Ο = ( b4 β a4) Γ β«2 cos2ΞΈ Γ sin2ΞΈ dΞΈ 0 4 2 2 = ( b4 β a4) Γ 1 Γ 1 Γ Ο (β΅ β« 2 2 Ο 2 0 1 1 Ο 4 2 2 cos ΞΈsin ΞΈdΞΈ = Γ Γ ) = Ο(b4βa4) 16 Example: π π Evaluate β« π β« βππβπ±π βππ β π±π β π²π ππ²ππ± by transforming into polar co-ordinates. Solution: Given order ππ¦ππ₯ is in correct form. Given limits are y : 0β βπ2 β π₯2 , x : 0β π Equations are y = 0 , y = βπ2 β π₯2 , x = 0 , x = π π¦2 = π2 β π₯2 x2 + y2 = a2 is a circle with centre (0,0) and radius βaβ. DOUBLE INTEGRATION
- 10. ENGG MATHS-II Replacement: Put a2 β x2 β y2 = a2 β (x2 + y2) = a2 β r2 , ππ¦ππ₯ = πππππ β΄ βa2 β x2 β y2 = βa2 β r2 Limits: r : 0 β π , π: 0 β π 2 π βπ2 β π2 πππππ π β« π β« βπ2βπ₯2 βπ2 β π₯2 β π¦2 ππ¦ππ₯ = β«2 β« 0 0 0 0 = 2 2 β π πππ)ππ π π β«2(β« βπ 0 0 Substitution: Put a2 β r2 = t β2rdr = dt if r = 0 β t = a2 if r = a β t = 0 rdr = β dt 2 β΄ t βΆ a2 β 0 2 2 a Ο β«2(β« βa β r 0 0 βdt 2 0 Ο rdr)dΞΈ = β«2[β« 2 βt( )]dΞΈ 0 a 2 β1 0 Ο = β«2[β« 2 βtdt]dΞΈ 0 a = β1 2 1 0 Ο β«2[β« 2 t2dt]dΞΈ 0 a = β1 2 β« t 3 2 3 2 a2 0 Ο 2 0 [ ] dΞΈ = β Γ 1 2 2 3 Ο 3 0 0 a2 β«2 [t2] dΞΈ 1 3 3 2 Ο = β β«2(0 β (a )2)dΞΈ = β 1 3 β« 0 Ο 3 β (a ) dΞΈ 2 0 = a3 3 Ο β«2 dΞΈ 0 a3 3 0 Ο = (ΞΈ)2 DOUBLE INTEGRATION
- 11. 3 2 = a3 ( Ο β 0) = Οa3 6 Exercise: Evaluate the following by changing into polar co-ordinates. 1. β« 2π β« β2π₯βπ₯2 ππ¦ππ₯ 0 0 Ans: ππ2 2 2. β« π β« βπ2βπ₯2 (π₯2 + π¦2)ππ¦ππ₯ 0 0 Ans: ππ4 8 π₯2 3. β« 1 β« 2βπ₯ π₯π¦ ππ₯ππ¦ 0 Ans: 3 8 π₯ ππ₯ππ¦ π¦ π₯2+π¦2 4. β« π β« π 0 Ans: ππ 4 π₯2+π¦2 ππ₯ππ¦ 5. β« 2π β« β2ππ₯βπ₯2 π₯ 0 0 Ans: ππ 2 6. β« π β« βπ2βπ₯2 (π₯2 + π¦2)ππ¦ππ₯ βπ 0 Ans: ππ4 4 7. β« π β« βπ2βπ₯2 (π₯2π¦ + π¦3)ππ₯ππ¦ 0 0 Ans: π5 5 8. β« 1 β« β2π₯βπ₯2 (π₯2 + π¦2)ππ¦ππ₯ 0 0 8 Ans: 3π β 1 X2+y2 9. β¬ X2y2 dxdy over the annular region between the circles x2 + y2 = 16 and x2 + y2 = 4 Ans:15Ο 10.β¬ Xy X2+y2 dxdy over the positive quadrant of the circle x2 + y2 = a2 Ans: a3 6 Change of Variables in Triple Integral Change of variables from Cartesian co- ordinates to cylindrical co β ordinates. To convert from Cartesian to cylindrical polar coordinates system we have the following transformation. π₯ = π πππ π π¦ = π π πππ π§ = π§ J = β(X ,y,z) = r β(r ,ΞΈ,z) Hence the integral becomes ENGG MATHS-II DOUBLE INTEGRATION
- 12. ENGG MATHS-II β f(x , y, z) dzdydx = β f(r , ΞΈ, z) dzdrdΞΈ Example: Find the volume of a solid bounded by the spherical surface π±π + π²π + π³π = πππ and the cylinder π±π + π²π β πππ² = π. Solution: Cylindrical co β ordinates x = r cos ΞΈ y = r sin ΞΈ z = z The equation of the sphere x2 + y2 + z2 = 4a2 r2cos2 ΞΈ + r2 sin2 ΞΈ + z2 = 4a2 r2 + z2 = 4a2 And the cylinder x2 + y2 β 2ay = 0 x2 + y2 = 2ay r2cos2 ΞΈ + r2 sin2 ΞΈ = 2a r sin ΞΈ r2 = 2arsin ΞΈ r = 2asin ΞΈ Hence, the required volume, DOUBLE INTEGRATION
- 13. β« β4a2β r2 r dz dr dΞΈ 0 rβ4a2 β r2 dr dΞΈ Volume = β« β« β« dx dy dz = β« β« β« r dΞΈ dr dz = 4 β« Οβ2 β« 2a sin ΞΈ 0 0 = 4 β« Οβ2 β« 2a sin ΞΈ 0 0 3 0 = 4 β« Οβ2 [β 1 (4a2 β r2)3β2] 2a sin ΞΈ dΞΈ 0 3 0 = 4 β« Οβ2 [β(4a2 β 4a2 sin2 ΞΈ)3β2 + 8a3] dΞΈ = 4 3 β« Οβ2 (β8a3 cos3 ΞΈ + 8a3) dΞΈ 0 3 = 4 8a3 β« Οβ2 (1 β cos3 ΞΈ) dΞΈ 0 3 2 3 = 32 a3 [ Ο β 2 ] cubic units Example: Find the volume of the portion of the cylinder π±π + π²π = π intercepted between the plane π± = π and the paraboloid π±π + π²π = π β π³. Solution: Cylindrical co β ordinates π₯ = π πππ π π¦ = π π ππ π π§ = π§ Given x2 + y2 = 1 r2cos2 ΞΈ + r2 sin2 ΞΈ = 1 r2 = 1 r = Β±1 Given x2 + y2 = 4 β z r2cos2 ΞΈ + r2 sin2 ΞΈ = 4 β z r2 = 4 β z z = 4 β r2 ENGG MATHS-II DOUBLE INTEGRATION
- 14. ENGG MATHS-II Hence the required volume Volume = β« β« β« r dz dr dΞΈ β« 4β r2 r dz dr dΞΈ 4β r 0 2 dr dΞΈ r (4 β r2) dr dΞΈ (4r β r3) dr dΞΈ = β« 2Ο β« 1 0 0 0 = β« 2Ο β« 1 r [z] 0 0 = β« 2Ο β« 1 0 0 = β« 2Ο β« 1 0 0 2 r4 β 4 ] dΞΈ 2Ο [4r2 = β«0 4 [(2 β 1 ) β (0 β 0)] dΞΈ = β« 2Ο 0 4 7 dΞΈ = β« 2Ο 0 4 0 = 7 [ΞΈ] 2Ο = 7 [2Ο β 0] 4 2 = 7 Ο cubic.units Example: Find the volume bounded by the paraboloid π±π + π²π = ππ³, and the cylinder π±π + π²π = πππ² and the plane π³ = π Solution: Cylindrical co β ordinates x = r cos ΞΈ y = r sin ΞΈ z = z The equation of the sphere x2 + y2+= az r2cos2 ΞΈ + r2 sin2 ΞΈ = az r2 = az And the cylinder x2 + y2 = 2ay r2cos2 ΞΈ + r2 sin2 ΞΈ = 2a r sin ΞΈ r2 = 2arsin ΞΈ DOUBLE INTEGRATION
- 15. r = 2asin ΞΈ Hence, the required volume, Volume = β« β« β« dx dy dz = β« β« β« r dΞΈ dr dz r2 0 Ο 2a sin ΞΈ =β« 0 β« 0 β« a r dz dr dΞΈ 0 r2 [z]a rdrdΞΈ β« 2a sin ΞΈ =β« Ο 0 0 r3 [ a ] drdΞΈ Ο 2a sin ΞΈ =β«0 β«0 = 1 β« a 4 0 r4 2a sin ΞΈ Ο 0 [ ] dΞΈ a 4 0 = 1 β« Ο 16a4sin4ΞΈ dΞΈ Ο 2 = 4a3 Γ 2 β« β sin4ΞΈdΞΈ 0 = 4a3 Γ 2 3 1 Ο = 3Οa3 4 2 2 2 Change of variables from Cartesian Co β ordinates to spherical Polar Co β ordinates To convert from Cartesian to spherical polar co-ordinates system we have the following transformation π₯ = ππ ππππππ π y = rsinΞΈsinΟ z = rcosΞΈ J = β(r ,ΞΈ,Ο ) β(X ,y,z) = r2sinΞΈ Hence the integral becomes β f(x , y, z) dzdydx = β f(r , ΞΈ, z)r2sinΞΈdrdΞΈdΟ Example: Evaluate β« β« β« π βπβ π±πβ π²πβ π³π ππ± ππ² ππ³ over the region bounded by the sphere π±π + π²π + π³π = π. Solution: Let us transform this integral in spherical polar co β ordinates by taking ENGG MATHS-II DOUBLE INTEGRATION
- 16. x = r sin ΞΈ cos Ο y = r sin ΞΈ sin Ο z = r cos ΞΈ dx dy dz = (r2 sin ΞΈ) dr dΞΈ dΟ Hence Ο varies from 0 to 2Ο Ο varies from 0 to Ο Ο varies from 0 to 1 r2 sin ΞΈ dr dΞΈ d Ο 0 β1β r2 = β« 2Ο β« Ο β« 1 1 0 0 0 0 r2 = [β« 2Ο dΟ ] [β« Ο sin ΞΈ dΞΈ] [β« 1 0 β1β r2 dr] 0 0 r2 dr = [Ο] 2Ο [β cosΞΈ] Ο β« 1 0 β1 β r2 r2 dr = (2Ο β 0) ( 1 + 1) β« 1 0 β1 β r2 r2 = 4Ο β« 1 dr 0 β1 β r2 Put r = sint ; dr = cost dt r = 0 β t = 0 r = 1 β t = Ο 2 sin2 t β1 β sin2t cost dt = 4Ο β« Οβ2 0 β cos2t = 4Ο β« Οβ2 sin2 t cost dt 0 cost = 4Ο β« Οβ2 sin2 t cos t dt 0 = 4Ο β« Οβ2 sin2 t dt 0 2 2 = 4Ο 1 Ο = Ο2 Example: β«βπ±π+π²π ππ³ ππ² ππ± βπ±π+ π²π+ π³π π βπβ π±π π Evaluate β«π β«π Solution: ENGG MATHS-II DOUBLE INTEGRATION
- 17. Given π₯ varies from 0 to 1 y varies from 0 to β1 β x2 z varies from βx2 + y2 to 1 Let us transform this integral into spherical polar co β ordinates by using x = r sin ΞΈ cos Ο y = r sin ΞΈ sin Ο z = r cos ΞΈ dx dy dz = (r2 sin ΞΈ) dr dΞΈ dΟ Let z = βx2 + y2 β z2 = x2 + y2 β r2 cos2 ΞΈ = r2 sin2ΞΈ cos2Ο + r2 sin2ΞΈ sin2Ο β cos2 ΞΈ = sin2 ΞΈ [β΅ cos2Ο + sin2Ο = 1 ] β ΞΈ = Ο 4 Let π§ = 1 β π πππ π = 1 β π = 1 πππ π β π = π ππ π The region of integration is common to the cone z2 = x2 + y2 and the cylinder x2 + y2 = 1 bounded by the plane z = 1 in the positive octant. r = 0 to r = sec ΞΈ ΞΈ = 0 to ΞΈ = Ο 4 Limits of r βΆ Limits of ΞΈ βΆ Limits of Ο βΆ Ο = 0 to Ο = Ο 2 r = β« Οβ2 β« Οβ4 β« sec ΞΈ 1 r2 sin ΞΈ dr dΞΈ dΟ 0 0 0 = β« Οβ2 β« Οβ4 β« sec ΞΈ r sin ΞΈ dr dΞΈ dΟ 0 0 0 r2 sec ΞΈ Οβ2 Οβ4 = β«0 β«0 [sin ΞΈ 2 ] 0 dΞΈ dΟ 2 Οβ2 Οβ4 sec2ΞΈ sin ΞΈβ0 = β«0 β«0 [ ] dΞΈ dΟ 2 0 0 = β« Οβ2 β« Οβ4 1 sec ΞΈ tan ΞΈ dΞΈ dΟ 2 0 0 = [1 β« Οβ2 dΟ ] [β« Οβ4 sec ΞΈ tan ΞΈ dΞΈ ] ENGG MATHS-II DOUBLE INTEGRATION
- 18. 2 0 0 = 1 [ ΞΈ ] Οβ2 [secΞΈ] Οβ4 2 2 = 1 [ Ο β 0] [β2 β 1] 4 = Ο (β2 β 1) Example: Evaluate β« β« β« (π±π + π²π + π³π )ππ± ππ² ππ³ taken over the region bounded by the volume enclosed by the sphere π±π + π²π + π³π = π. Solution: Let us convert the given integral into spherical polar co β ordinates. x = r sin ΞΈ cos Ο β x2 = r2 sin2 ΞΈ cos2 Ο y = r sin ΞΈ sin Ο β y2 = r2 sin2 ΞΈ sin2 Ο z = r cos ΞΈ β z2 = r2 cos2 ΞΈ dx dy dz = (r2 sin ΞΈ) dr dΞΈ dΟ β« β« β« (x2 + y2 + z2 )dx dy dz = β« Ο β« 2Ο β« 1 r2 (r2 sin ΞΈ dΞΈ dΟ dr) 0 0 0 Limits of r βΆ Limits of ΞΈ βΆ Limits of Ο βΆ r = 0 to r = 1 ΞΈ = 0 to ΞΈ = Ο Ο = 0 to Ο = 2Ο β« β« β« (x2 + y2 + z2 )dx dy dz = β« Ο β« 2Ο β« 1 r2 (r2 sin ΞΈ dΞΈ dΟ dr) 0 0 0 = [β« 1 r4 dr] [β« Ο sin ΞΈ dΞΈ] [β« 2Ο dΟ] 0 0 0 r5 1 0 Ο = [ 5 ] 0 [β cos ΞΈ] [Ο]2Ο 0 5 = ( 1 β 0) (1 + 1) (2Ο β 0) = ( 1 ) (2) (2Ο) = 4Ο 5 5 ENGG MATHS-II DOUBLE INTEGRATION