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Double Integration examples of double integration with substitution.pptx

The document provides examples of evaluating double integrals using polar coordinates. It discusses: 1) Converting Cartesian coordinates (x,y) to polar coordinates (r,θ) and setting up the integral limits and differentials correctly based on the region of integration. 2) Evaluating the integrals using properties of trigonometric functions like cos2θ, cos4θ, and the integral of secθ. 3) Examples include integrals over circular and annular regions.

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Change into polar co-ordinates and then evaluate ∫ ∞ ∫ ∞ 𝐞−(𝐱𝟐+𝐲𝟐)𝐝𝐲𝐝𝐱 𝟎 𝟎 Solution: Given order 𝑑𝑦𝑑𝑥 is in correct form. Given limits are 𝑦 ∶ 0 → ∞ , 𝑥 ∶ 0 → ∞ Equations are 𝑦 = 0 , 𝑦 = ∞ , 𝑥 = 0 , 𝑥 = ∞ 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 = rcos𝜃 , y = rsin𝜃 , x2 + y2 = r2 and dxdy = rdrd𝜃 Step:5 Find r limits(draw radial strip inside the region) and θ limits and evaluate the integral. Example: DOUBLE INTEGRATION
ENGG MATHS-II Put x2 + y2 = r2 , dydx = rdrd𝜃 Limits: r : 0→ ∞ , θ: 0 → π 2 ∞ e−r2 rdrdθ π ∫ ∞ ∫ ∞ e−(X2+y2)dydx = ∫2 ∫ 0 0 0 0 Substitution: Put r2 = t , if r = 0 ⇒ t = 0, r = ∞ ⇒ t = ∞ 2rdr = dt t: 0→ ∞ rdr = 𝑑𝑡 π ∫2 ∫ 0 0 2 ∞ e−t dt dθ π 2 ∞ e−r2 rdrdθ = ∫2 ∫ 0 0 1 2 −1 0 e−t ∞ π 0 = ∫2 [ ] dθ = −∞ 0 + e )dθ π ∫2(−e 0 = 1 2 1 2 π ∫2(0 + 1)dθ 0 (∵ e−∞ = 0, e0 = 1) 1 2 π = ∫2 dθ 0 2 0 π = 1 (θ)2 = 1 ( π − 0) = 2 2 π 4 Example: 𝐱 𝐝𝐲𝐝𝐱 𝐲 𝐱𝟐+𝐲𝟐 Change into polar co-ordinates and then evaluate ∫ 𝐚 ∫ 𝐚 𝟎 Solution: Given order 𝑑𝑥𝑑𝑦 is in correct form. DOUBLE INTEGRATION
ENGG MATHS-II Given limits are x : y→ 𝑎 , y : 0→ 𝑎 Equations are x =y , x = 𝑎 , y = 0 , y = 𝑎 Replacement: Put x = rcos𝜃 , 𝑥2 + 𝑦2 = 𝑟2 , 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑𝜃 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 𝑐𝑜𝑠𝜃 4 ∫ ∫ 𝑎 𝑎 𝑥 𝑥 +𝑦 2 2 𝑑𝑦𝑑𝑥 = ∫ ∫ 0 𝑦 𝑟2 𝑟𝑑𝑟𝑑𝜃 𝑎 𝑟𝑐𝑜𝑠𝜃 𝜋 4 𝑐𝑜 𝑠 𝜃 0 0 0 𝑎 𝜋 = ∫4[𝑟𝑐𝑜𝑠𝜃]𝑐𝑜𝑠𝜃 𝑑𝜃 𝑎 𝑐𝑜𝑠 𝜃 0 𝜋 = ∫4 ( 0 𝑐𝑜𝑠𝜃 − 0) 𝑑𝜃 𝜋 = 𝑎 ∫4 𝑑𝜃 0 0 𝜋 = 𝑎(𝜃)4 4 = 𝑎( 𝜋 − 0) = 𝑎𝜋 4 Example: 𝐱𝟐 𝐲 √𝐱𝟐+𝐲𝟐 Evaluate ∫ 𝐚 ∫ 𝐚 𝟎 𝐝𝐱𝐝𝐲 by changing into polar co-ordinates. Solution: Given order 𝑑𝑥𝑑𝑦 is in correct form. Given limits are x : y→ 𝑎 , y : 0→ 𝑎 DOUBLE INTEGRATION
ENGG MATHS-II Equations are x =y , x = 𝑎 , y = 0 , y = 𝑎 Replacement: Put x2 = r2cos2𝜃 , 𝑥2 + 𝑦2 = 𝑟2 => 𝑟 = √𝑥2 + 𝑦2 , dxdy = rdrd𝜃 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 𝑐𝑜𝑠𝜃 4 ∫ ∫ 𝑥2 √𝑥 +𝑦 2 2 𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑎 𝑎 0 𝑦 𝑟 𝑟𝑑𝑟𝑑𝜃 𝑎 𝑟2𝑐𝑜𝑠2𝜃 𝜋 4 𝑐 𝑜 𝑠 𝜃 0 0 3 3 𝜋 = ∫4 [𝑟 𝑎 𝑐𝑜𝑠2𝜃] 𝑐𝑜𝑠𝜃 𝑑𝜃 0 𝑎3 3 3𝑐𝑜𝑠 𝜃 2 0 𝜋 = ∫4 ( 0 𝑐𝑜𝑠 𝜃 − 0) 𝑑𝜃 = 𝑎3 1 3 2 𝜋 ∫4 3 0 𝑐𝑜𝑠 𝜃 𝑐𝑜𝑠 𝜃 𝑑𝜃 = 𝑎3 1 𝜋 3 0 𝑐𝑜𝑠𝜃 ∫4 𝑑𝜃 = 𝜋 0 ∫4 𝑠𝑒𝑐𝜃 𝑑𝜃 = 𝑎3 3 𝑎3 3 0 𝜋 (𝑙𝑜𝑔(𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃))4 = 𝑎3 [𝑙𝑜𝑔 (𝑠𝑒𝑐 𝜋 + 𝑡𝑎𝑛 𝜋 ) − 𝑙𝑜𝑔(𝑠𝑒𝑐0 + 𝑡𝑎𝑛0)] 3 4 4 3 = 𝑎3 [𝑙𝑜𝑔(√2 + 1) − 𝑙𝑜𝑔(1 − 0)] 3 = 𝑎3 𝑙𝑜𝑔(√2 + 1) Note: 1. x2 + y2 = r2cos2θ + r2sin2θ = r2 2. ∫ cos 2 0 θdθ = ∫ 2 2 π π 2 0 sin θdθ = × 1 π 2 2 3. ∫ 4 cos θdθ = ∫ π 2 0 4 π 2 0 sin θdθ = × × 3 1 π 4 2 2 4. ∫ 2 2 π 2 0 cos θsin θdθ = × × 1 1 π 4 2 2 DOUBLE INTEGRATION
Example: By changing into polar co-ordinates and evaluate ∫ 𝟐𝐚 ∫ √𝟐𝐚𝐱−𝐱𝟐 (𝐱𝟐 + 𝐲𝟐)𝐝𝐲𝐝𝐱 𝟎 𝟎 Solution: Given order 𝑑𝑦𝑑𝑥 is in correct form. Given limits are y : 0→ √2𝑎𝑥 − 𝑥2 , x : 0→ 2𝑎 Equations are y = 0 , y = √2𝑎𝑥 − 𝑥2 , x = 0 , x = 2𝑎 y2 = 2ax − x2 x2 + y2 − 2ax = 0 is a circle with centre (a,0) and radius ‘a’. Replacement: Put x2 + y2 = r2 , 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑𝜃 Limits: r : 0 → 2𝑎𝑐𝑜𝑠𝜃 , 𝜃: 0 → 𝜋 2 2𝑎𝑐𝑜𝑠𝜃 𝑟2 × 𝑟𝑑𝑟𝑑𝜃 𝜋 ∫ 2𝑎 ∫ √2𝑎𝑥−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 = ∫2 ∫ 0 0 0 0 3 𝑟 𝑑𝑟𝑑𝜃 2𝑎𝑐𝑜𝑠 𝜃 𝜋 = ∫2 ∫ 0 0 4 ] 0 r4 2acosθ π = ∫2 [ 0 dθ 4 π 24a4cos4θ = ∫2( 0 − 0) dθ π 0 = 4a4 ∫2 cos4θdθ ENGG MATHS-II DOUBLE INTEGRATION
= 4a4 × 3 × 1 × π 4 2 2 (∵ ∫ 4 π 2 0 cos θdθ = × × 3 1 π 4 2 2 ) = 3πa4 4 Example: 𝐱 𝐱𝟐+𝐲𝟐 𝐝𝐱𝐝𝐲 By changing into polar co-ordinates and evaluate∫ 𝟐 ∫ √𝟐𝐱−𝐱𝟐 𝟎 𝟎 Solution: Given order 𝑑𝑥𝑑𝑦 is in incorrect form. X X2+y2 dydx The correct form is 𝑑𝑦𝑑𝑥 ⇒ ∫ 2 ∫ √2X−X2 0 0 Given limits are y : 0→ √2𝑥 − 𝑥2 , x : 0→ 2 Equations are y = 0 , y = √2𝑥 − 𝑥2 , x = 0 , x = 2 y2 = 2x − x2 x2 + y2 − 2x = 0 is a circle with centre (1,0) and radius ‘1’. Replacement: Put x = rcos𝜃, 𝑥2 + 𝑦2 = 𝑟2 , dxdy = rdrd𝜃 Limits: r : 0 → 2𝑐𝑜𝑠𝜃 , 𝜃: 0 → 𝜋 2 𝑥 +𝑦 2 2 𝑑𝑦𝑑𝑥 = ∫ ∫ 2 √2𝑥−𝑥2 𝑥 ∫0∫0 𝑟2 2𝑐𝑜𝑠𝜃 𝑟𝑐𝑜𝑠𝜃 × 𝑟𝑑𝑟𝑑𝜃 𝜋 2 0 0 0 𝜋 0 =∫2[𝑟𝑐𝑜𝑠𝜃]2𝑐𝑜𝑠𝜃 𝑑𝜃 𝜋 0 = ∫2(2𝑐𝑜𝑠2𝜃 − 0)𝑑𝜃 ENGG MATHS-II DOUBLE INTEGRATION
𝜋 0 = 2 ∫2 𝑐𝑜𝑠2𝜃 𝑑𝜃 = 2 × 1 × 𝜋 2 2 (∵ ∫ 2 𝜋 2 0 1 𝜋 2 2 𝑐𝑜𝑠 𝜃𝑑𝜃 = × ) = 𝜋 2 Example: By changing into polar co-ordinates and evaluate∫ 𝐚 ∫ √𝐚𝟐−𝐱𝟐 √𝐱𝟐 + 𝐲𝟐𝐝𝐲𝐝𝐱 𝟎 𝟎 Solution: Given order 𝑑𝑥𝑑𝑦 is in correct form. Given limits are y : 0→ √𝑎2 − 𝑥2 , x : 0→ 𝑎 Equations are y = 0 , y = √𝑎2 − 𝑥2 , x = 0 , x = 𝑎 y2 = a2 − x2 x2 + y2 = a2 is a circle with centre (0,0) and radius ‘a’. Replacement: Put x2 + y2 = r2 ⇒ r = √x2 + y2 , 𝑑𝑦𝑑𝑥 = 𝑟𝑑𝑟𝑑𝜃 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 a r × rdrdθ 2 π ∫ a ∫ √a2−X2 √x2 + y2dydx = ∫2 ∫ 0 0 0 0 a r2drdθ π = ∫2 ∫ 0 0 3 π r3 a = ∫2 [ 0 ] dθ 0 ENGG MATHS-II DOUBLE INTEGRATION
a3 3 π = ∫2 ( 0 − 0)dθ = a3 3 π ∫2 dθ 0 a3 3 0 π = (θ)2 3 2 = a3 ( π − 0) = πa3 6 Example: 𝐱𝟐+𝐲𝟐 Evaluate ∬ 𝐱𝟐𝐲𝟐 𝐝𝐱𝐝𝐲 over the annular region between the circles x2 + y2 = a2 and x2 + y2 = b2 (b>a) by transforming into polar co-ordinates. Solution: 𝑥2+𝑦2 𝑟2 Replacement: Put x2 = r2cos2θ,y2 = r2sin2θ x2 + y2 = r2, 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑𝜃 Given the region is between the circles x2 + y2 = a2 and x2 + y2 = b2 Limits: r : 𝑎 → 𝑏 , 𝜃: 0 → 2𝜋 ∴ ∬ 𝑥2𝑦2 𝑑𝑥𝑑𝑦 = ∫ 2𝜋 ∫ 𝑏 𝑟2𝑐𝑜𝑠2𝜃×𝑟2𝑠𝑖𝑛2𝜃 × 𝑟𝑑𝑟𝑑𝜃 0 𝑎 ENGG MATHS-II DOUBLE INTEGRATION
r2 0 a ENGG MATHS-II = ∫ 2π ∫ b r5cos2θ×sin2θ × drdθ = ∫ 2π ∫ b r3cos2θ × sin2θ drdθ 0 a 4 a r4 b 2π 2 2 = ∫0 [ ] cos θ × sin θ dθ 4 0 = 1 ∫ 2π (b4 − a4) cos2θ × sin2θ dθ 4 = (b4−a4) ∫ 2π cos2θ × sin2θ dθ 0 = 4 4 (b −a ) 4 × ∫ π 2 4 0 2π 0 cos2θ × sin2θ dθ (∵ ∫ = 4∫ ) π 2 0 π = ( b4 − a4) × ∫2 cos2θ × sin2θ dθ 0 4 2 2 = ( b4 − a4) × 1 × 1 × π (∵ ∫ 2 2 π 2 0 1 1 π 4 2 2 cos θsin θdθ = × × ) = π(b4−a4) 16 Example: 𝟎 𝟎 Evaluate ∫ 𝐚 ∫ √𝐚𝟐−𝐱𝟐 √𝐚𝟐 − 𝐱𝟐 − 𝐲𝟐 𝐝𝐲𝐝𝐱 by transforming into polar co-ordinates. Solution: Given order 𝑑𝑦𝑑𝑥 is in correct form. Given limits are y : 0→ √𝑎2 − 𝑥2 , x : 0→ 𝑎 Equations are y = 0 , y = √𝑎2 − 𝑥2 , x = 0 , x = 𝑎 𝑦2 = 𝑎2 − 𝑥2 x2 + y2 = a2 is a circle with centre (0,0) and radius ‘a’. DOUBLE INTEGRATION
ENGG MATHS-II Replacement: Put a2 − x2 − y2 = a2 − (x2 + y2) = a2 − r2 , 𝑑𝑦𝑑𝑥 = 𝑟𝑑𝑟𝑑𝜃 ∴ √a2 − x2 − y2 = √a2 − r2 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 2 𝑎 √𝑎2 − 𝑟2 𝑟𝑑𝑟𝑑𝜃 𝜋 ∫ 𝑎 ∫ √𝑎2−𝑥2 √𝑎2 − 𝑥2 − 𝑦2 𝑑𝑦𝑑𝑥 = ∫2 ∫ 0 0 0 0 = 2 2 − 𝑟 𝑟𝑑𝑟)𝑑𝜃 𝑎 𝜋 ∫2(∫ √𝑎 0 0 Substitution: Put a2 − r2 = t −2rdr = dt if r = 0 ⇒ t = a2 if r = a ⇒ t = 0 rdr = − dt 2 ∴ t ∶ a2 → 0 2 2 a π ∫2(∫ √a − r 0 0 −dt 2 0 π rdr)dθ = ∫2[∫ 2 √t( )]dθ 0 a 2 −1 0 π = ∫2[∫ 2 √tdt]dθ 0 a = −1 2 1 0 π ∫2[∫ 2 t2dt]dθ 0 a = −1 2 ∫ t 3 2 3 2 a2 0 π 2 0 [ ] dθ = − × 1 2 2 3 π 3 0 0 a2 ∫2 [t2] dθ 1 3 3 2 π = − ∫2(0 − (a )2)dθ = − 1 3 ∫ 0 π 3 − (a ) dθ 2 0 = a3 3 π ∫2 dθ 0 a3 3 0 π = (θ)2 DOUBLE INTEGRATION
3 2 = a3 ( π − 0) = πa3 6 Exercise: Evaluate the following by changing into polar co-ordinates. 1. ∫ 2𝑎 ∫ √2𝑥−𝑥2 𝑑𝑦𝑑𝑥 0 0 Ans: 𝜋𝑎2 2 2. ∫ 𝑎 ∫ √𝑎2−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 0 0 Ans: 𝜋𝑎4 8 𝑥2 3. ∫ 1 ∫ 2−𝑥 𝑥𝑦 𝑑𝑥𝑑𝑦 0 Ans: 3 8 𝑥 𝑑𝑥𝑑𝑦 𝑦 𝑥2+𝑦2 4. ∫ 𝑎 ∫ 𝑎 0 Ans: 𝜋𝑎 4 𝑥2+𝑦2 𝑑𝑥𝑑𝑦 5. ∫ 2𝑎 ∫ √2𝑎𝑥−𝑥2 𝑥 0 0 Ans: 𝜋𝑎 2 6. ∫ 𝑎 ∫ √𝑎2−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 −𝑎 0 Ans: 𝜋𝑎4 4 7. ∫ 𝑎 ∫ √𝑎2−𝑥2 (𝑥2𝑦 + 𝑦3)𝑑𝑥𝑑𝑦 0 0 Ans: 𝑎5 5 8. ∫ 1 ∫ √2𝑥−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 0 0 8 Ans: 3𝜋 − 1 X2+y2 9. ∬ X2y2 dxdy over the annular region between the circles x2 + y2 = 16 and x2 + y2 = 4 Ans:15π 10.∬ 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 – ordinates. To convert from Cartesian to cylindrical polar coordinates system we have the following transformation. 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑦 = 𝑟 𝑠𝑖𝑛𝜃 𝑧 = 𝑧 J = ∂(X ,y,z) = r ∂(r ,θ,z) Hence the integral becomes ENGG MATHS-II DOUBLE INTEGRATION
ENGG MATHS-II ∭ f(x , y, z) dzdydx = ∭ f(r , θ, z) dzdrdθ Example: Find the volume of a solid bounded by the spherical surface 𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 = 𝟒𝐚𝟐 and the cylinder 𝐱𝟐 + 𝐲𝟐 − 𝟐𝐚𝐲 = 𝟎. Solution: Cylindrical co – ordinates x = r cos θ y = r sin θ z = z The equation of the sphere x2 + y2 + z2 = 4a2 r2cos2 θ + r2 sin2 θ + z2 = 4a2 r2 + z2 = 4a2 And the cylinder x2 + y2 − 2ay = 0 x2 + y2 = 2ay r2cos2 θ + r2 sin2 θ = 2a r sin θ r2 = 2arsin θ r = 2asin θ Hence, the required volume, DOUBLE INTEGRATION
∫ √4a2− r2 r dz dr dθ 0 r√4a2 − r2 dr dθ Volume = ∫ ∫ ∫ dx dy dz = ∫ ∫ ∫ r dθ dr dz = 4 ∫ π⁄2 ∫ 2a sin θ 0 0 = 4 ∫ π⁄2 ∫ 2a sin θ 0 0 3 0 = 4 ∫ π⁄2 [− 1 (4a2 − r2)3⁄2] 2a sin θ dθ 0 3 0 = 4 ∫ π⁄2 [−(4a2 − 4a2 sin2 θ)3⁄2 + 8a3] dθ = 4 3 ∫ π⁄2 (−8a3 cos3 θ + 8a3) dθ 0 3 = 4 8a3 ∫ π⁄2 (1 − cos3 θ) dθ 0 3 2 3 = 32 a3 [ π − 2 ] cubic units Example: Find the volume of the portion of the cylinder 𝐱𝟐 + 𝐲𝟐 = 𝟏 intercepted between the plane 𝐱 = 𝟎 and the paraboloid 𝐱𝟐 + 𝐲𝟐 = 𝟒 − 𝐳. Solution: Cylindrical co – ordinates 𝑥 = 𝑟 𝑐𝑜𝑠 𝜃 𝑦 = 𝑟 𝑠𝑖𝑛 𝜃 𝑧 = 𝑧 Given x2 + y2 = 1 r2cos2 θ + r2 sin2 θ = 1 r2 = 1 r = ±1 Given x2 + y2 = 4 − z r2cos2 θ + r2 sin2 θ = 4 − z r2 = 4 − z z = 4 − r2 ENGG MATHS-II DOUBLE INTEGRATION
ENGG MATHS-II Hence the required volume Volume = ∫ ∫ ∫ r dz dr dθ ∫ 4− r2 r dz dr dθ 4− r 0 2 dr dθ r (4 − r2) dr dθ (4r − r3) dr dθ = ∫ 2π ∫ 1 0 0 0 = ∫ 2π ∫ 1 r [z] 0 0 = ∫ 2π ∫ 1 0 0 = ∫ 2π ∫ 1 0 0 2 r4 − 4 ] dθ 2π [4r2 = ∫0 4 [(2 − 1 ) − (0 − 0)] dθ = ∫ 2π 0 4 7 dθ = ∫ 2π 0 4 0 = 7 [θ] 2π = 7 [2π − 0] 4 2 = 7 π cubic.units Example: Find the volume bounded by the paraboloid 𝐱𝟐 + 𝐲𝟐 = 𝐚𝐳, and the cylinder 𝐱𝟐 + 𝐲𝟐 = 𝟐𝐚𝐲 and the plane 𝐳 = 𝟎 Solution: Cylindrical co – ordinates x = r cos θ y = r sin θ z = z The equation of the sphere x2 + y2+= az r2cos2 θ + r2 sin2 θ = az r2 = az And the cylinder x2 + y2 = 2ay r2cos2 θ + r2 sin2 θ = 2a r sin θ r2 = 2arsin θ DOUBLE INTEGRATION
r = 2asin θ Hence, the required volume, Volume = ∫ ∫ ∫ dx dy dz = ∫ ∫ ∫ r dθ dr dz r2 0 π 2a sin θ =∫ 0 ∫ 0 ∫ a r dz dr dθ 0 r2 [z]a rdrdθ ∫ 2a sin θ =∫ π 0 0 r3 [ a ] drdθ π 2a sin θ =∫0 ∫0 = 1 ∫ a 4 0 r4 2a sin θ π 0 [ ] dθ a 4 0 = 1 ∫ π 16a4sin4θ dθ π 2 = 4a3 × 2 ∫ ⁄ sin4θdθ 0 = 4a3 × 2 3 1 π = 3πa3 4 2 2 2 Change of variables from Cartesian Co – ordinates to spherical Polar Co – ordinates To convert from Cartesian to spherical polar co-ordinates system we have the following transformation 𝑥 = 𝑟𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜑 y = rsinθsinφ z = rcosθ J = ∂(r ,θ,φ ) ∂(X ,y,z) = r2sinθ Hence the integral becomes ∭ f(x , y, z) dzdydx = ∭ f(r , θ, z)r2sinθdrdθdφ Example: Evaluate ∫ ∫ ∫ 𝟏 √𝟏− 𝐱𝟐− 𝐲𝟐− 𝐳𝟐 𝐝𝐱 𝐝𝐲 𝐝𝐳 over the region bounded by the sphere 𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 = 𝟏. Solution: Let us transform this integral in spherical polar co – ordinates by taking ENGG MATHS-II DOUBLE INTEGRATION
x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ dx dy dz = (r2 sin θ) dr dθ dϕ Hence ϕ varies from 0 to 2π ϕ varies from 0 to π ϕ varies from 0 to 1 r2 sin θ dr dθ d ϕ 0 √1− r2 = ∫ 2π ∫ π ∫ 1 1 0 0 0 0 r2 = [∫ 2π dϕ ] [∫ π sin θ dθ] [∫ 1 0 √1− r2 dr] 0 0 r2 dr = [ϕ] 2π [– cosθ] π ∫ 1 0 √1 − r2 r2 dr = (2π − 0) ( 1 + 1) ∫ 1 0 √1 − r2 r2 = 4π ∫ 1 dr 0 √1 − r2 Put r = sint ; dr = cost dt r = 0 ⇒ t = 0 r = 1 ⇒ t = π 2 sin2 t √1 − sin2t cost dt = 4π ∫ π⁄2 0 √ cos2t = 4π ∫ π⁄2 sin2 t cost dt 0 cost = 4π ∫ π⁄2 sin2 t cos t dt 0 = 4π ∫ π⁄2 sin2 t dt 0 2 2 = 4π 1 π = π2 Example: ∫√𝐱𝟐+𝐲𝟐 𝐝𝐳 𝐝𝐲 𝐝𝐱 √𝐱𝟐+ 𝐲𝟐+ 𝐳𝟐 𝟏 √𝟏− 𝐱𝟐 𝟏 Evaluate ∫𝟎 ∫𝟎 Solution: ENGG MATHS-II DOUBLE INTEGRATION
Given 𝑥 varies from 0 to 1 y varies from 0 to √1 − x2 z varies from √x2 + y2 to 1 Let us transform this integral into spherical polar co – ordinates by using x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ dx dy dz = (r2 sin θ) dr dθ dϕ Let z = √x2 + y2 ⇒ z2 = x2 + y2 ⇒ r2 cos2 θ = r2 sin2θ cos2ϕ + r2 sin2θ sin2ϕ ⇒ cos2 θ = sin2 θ [∵ cos2ϕ + sin2ϕ = 1 ] ⇒ θ = π 4 Let 𝑧 = 1 ⇒ 𝑟 𝑐𝑜𝑠 𝜃 = 1 ⇒ 𝑟 = 1 𝑐𝑜𝑠 𝜃 ⇒ 𝑟 = 𝑠𝑒𝑐 𝜃 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 θ θ = 0 to θ = π 4 Limits of r ∶ Limits of θ ∶ Limits of ϕ ∶ ϕ = 0 to ϕ = π 2 r = ∫ π⁄2 ∫ π⁄4 ∫ sec θ 1 r2 sin θ dr dθ dϕ 0 0 0 = ∫ π⁄2 ∫ π⁄4 ∫ sec θ r sin θ dr dθ dϕ 0 0 0 r2 sec θ π⁄2 π⁄4 = ∫0 ∫0 [sin θ 2 ] 0 dθ dϕ 2 π⁄2 π⁄4 sec2θ sin θ−0 = ∫0 ∫0 [ ] dθ dϕ 2 0 0 = ∫ π⁄2 ∫ π⁄4 1 sec θ tan θ dθ dϕ 2 0 0 = [1 ∫ π⁄2 dϕ ] [∫ π⁄4 sec θ tan θ dθ ] ENGG MATHS-II DOUBLE INTEGRATION
2 0 0 = 1 [ θ ] π⁄2 [secθ] π⁄4 2 2 = 1 [ π − 0] [√2 − 1] 4 = π (√2 − 1) Example: Evaluate ∫ ∫ ∫ (𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 )𝐝𝐱 𝐝𝐲 𝐝𝐳 taken over the region bounded by the volume enclosed by the sphere 𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 = 𝟏. Solution: Let us convert the given integral into spherical polar co – ordinates. x = r sin θ cos ϕ ⇒ x2 = r2 sin2 θ cos2 ϕ y = r sin θ sin ϕ ⇒ y2 = r2 sin2 θ sin2 ϕ z = r cos θ ⇒ z2 = r2 cos2 θ dx dy dz = (r2 sin θ) dr dθ dϕ ∫ ∫ ∫ (x2 + y2 + z2 )dx dy dz = ∫ π ∫ 2π ∫ 1 r2 (r2 sin θ dθ dϕ dr) 0 0 0 Limits of r ∶ Limits of θ ∶ Limits of ϕ ∶ r = 0 to r = 1 θ = 0 to θ = π ϕ = 0 to ϕ = 2π ∫ ∫ ∫ (x2 + y2 + z2 )dx dy dz = ∫ π ∫ 2π ∫ 1 r2 (r2 sin θ dθ dϕ dr) 0 0 0 = [∫ 1 r4 dr] [∫ π sin θ dθ] [∫ 2π dϕ] 0 0 0 r5 1 0 π = [ 5 ] 0 [– cos θ] [ϕ]2π 0 5 = ( 1 − 0) (1 + 1) (2π − 0) = ( 1 ) (2) (2π) = 4π 5 5 ENGG MATHS-II DOUBLE INTEGRATION

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Double Integration examples of double integration with substitution.pptx

  • 1.
    Change into polarco-ordinates and then evaluate ∫ ∞ ∫ ∞ 𝐞−(𝐱𝟐+𝐲𝟐)𝐝𝐲𝐝𝐱 𝟎 𝟎 Solution: Given order 𝑑𝑦𝑑𝑥 is in correct form. Given limits are 𝑦 ∶ 0 → ∞ , 𝑥 ∶ 0 → ∞ Equations are 𝑦 = 0 , 𝑦 = ∞ , 𝑥 = 0 , 𝑥 = ∞ 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 = rcos𝜃 , y = rsin𝜃 , x2 + y2 = r2 and dxdy = rdrd𝜃 Step:5 Find r limits(draw radial strip inside the region) and θ limits and evaluate the integral. Example: DOUBLE INTEGRATION
  • 2.
    ENGG MATHS-II Put x2+ y2 = r2 , dydx = rdrd𝜃 Limits: r : 0→ ∞ , θ: 0 → π 2 ∞ e−r2 rdrdθ π ∫ ∞ ∫ ∞ e−(X2+y2)dydx = ∫2 ∫ 0 0 0 0 Substitution: Put r2 = t , if r = 0 ⇒ t = 0, r = ∞ ⇒ t = ∞ 2rdr = dt t: 0→ ∞ rdr = 𝑑𝑡 π ∫2 ∫ 0 0 2 ∞ e−t dt dθ π 2 ∞ e−r2 rdrdθ = ∫2 ∫ 0 0 1 2 −1 0 e−t ∞ π 0 = ∫2 [ ] dθ = −∞ 0 + e )dθ π ∫2(−e 0 = 1 2 1 2 π ∫2(0 + 1)dθ 0 (∵ e−∞ = 0, e0 = 1) 1 2 π = ∫2 dθ 0 2 0 π = 1 (θ)2 = 1 ( π − 0) = 2 2 π 4 Example: 𝐱 𝐝𝐲𝐝𝐱 𝐲 𝐱𝟐+𝐲𝟐 Change into polar co-ordinates and then evaluate ∫ 𝐚 ∫ 𝐚 𝟎 Solution: Given order 𝑑𝑥𝑑𝑦 is in correct form. DOUBLE INTEGRATION
  • 3.
    ENGG MATHS-II Given limitsare x : y→ 𝑎 , y : 0→ 𝑎 Equations are x =y , x = 𝑎 , y = 0 , y = 𝑎 Replacement: Put x = rcos𝜃 , 𝑥2 + 𝑦2 = 𝑟2 , 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑𝜃 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 𝑐𝑜𝑠𝜃 4 ∫ ∫ 𝑎 𝑎 𝑥 𝑥 +𝑦 2 2 𝑑𝑦𝑑𝑥 = ∫ ∫ 0 𝑦 𝑟2 𝑟𝑑𝑟𝑑𝜃 𝑎 𝑟𝑐𝑜𝑠𝜃 𝜋 4 𝑐𝑜 𝑠 𝜃 0 0 0 𝑎 𝜋 = ∫4[𝑟𝑐𝑜𝑠𝜃]𝑐𝑜𝑠𝜃 𝑑𝜃 𝑎 𝑐𝑜𝑠 𝜃 0 𝜋 = ∫4 ( 0 𝑐𝑜𝑠𝜃 − 0) 𝑑𝜃 𝜋 = 𝑎 ∫4 𝑑𝜃 0 0 𝜋 = 𝑎(𝜃)4 4 = 𝑎( 𝜋 − 0) = 𝑎𝜋 4 Example: 𝐱𝟐 𝐲 √𝐱𝟐+𝐲𝟐 Evaluate ∫ 𝐚 ∫ 𝐚 𝟎 𝐝𝐱𝐝𝐲 by changing into polar co-ordinates. Solution: Given order 𝑑𝑥𝑑𝑦 is in correct form. Given limits are x : y→ 𝑎 , y : 0→ 𝑎 DOUBLE INTEGRATION
  • 4.
    ENGG MATHS-II Equations arex =y , x = 𝑎 , y = 0 , y = 𝑎 Replacement: Put x2 = r2cos2𝜃 , 𝑥2 + 𝑦2 = 𝑟2 => 𝑟 = √𝑥2 + 𝑦2 , dxdy = rdrd𝜃 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 𝑐𝑜𝑠𝜃 4 ∫ ∫ 𝑥2 √𝑥 +𝑦 2 2 𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑎 𝑎 0 𝑦 𝑟 𝑟𝑑𝑟𝑑𝜃 𝑎 𝑟2𝑐𝑜𝑠2𝜃 𝜋 4 𝑐 𝑜 𝑠 𝜃 0 0 3 3 𝜋 = ∫4 [𝑟 𝑎 𝑐𝑜𝑠2𝜃] 𝑐𝑜𝑠𝜃 𝑑𝜃 0 𝑎3 3 3𝑐𝑜𝑠 𝜃 2 0 𝜋 = ∫4 ( 0 𝑐𝑜𝑠 𝜃 − 0) 𝑑𝜃 = 𝑎3 1 3 2 𝜋 ∫4 3 0 𝑐𝑜𝑠 𝜃 𝑐𝑜𝑠 𝜃 𝑑𝜃 = 𝑎3 1 𝜋 3 0 𝑐𝑜𝑠𝜃 ∫4 𝑑𝜃 = 𝜋 0 ∫4 𝑠𝑒𝑐𝜃 𝑑𝜃 = 𝑎3 3 𝑎3 3 0 𝜋 (𝑙𝑜𝑔(𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃))4 = 𝑎3 [𝑙𝑜𝑔 (𝑠𝑒𝑐 𝜋 + 𝑡𝑎𝑛 𝜋 ) − 𝑙𝑜𝑔(𝑠𝑒𝑐0 + 𝑡𝑎𝑛0)] 3 4 4 3 = 𝑎3 [𝑙𝑜𝑔(√2 + 1) − 𝑙𝑜𝑔(1 − 0)] 3 = 𝑎3 𝑙𝑜𝑔(√2 + 1) Note: 1. x2 + y2 = r2cos2θ + r2sin2θ = r2 2. ∫ cos 2 0 θdθ = ∫ 2 2 π π 2 0 sin θdθ = × 1 π 2 2 3. ∫ 4 cos θdθ = ∫ π 2 0 4 π 2 0 sin θdθ = × × 3 1 π 4 2 2 4. ∫ 2 2 π 2 0 cos θsin θdθ = × × 1 1 π 4 2 2 DOUBLE INTEGRATION
  • 5.
    Example: By changing intopolar co-ordinates and evaluate ∫ 𝟐𝐚 ∫ √𝟐𝐚𝐱−𝐱𝟐 (𝐱𝟐 + 𝐲𝟐)𝐝𝐲𝐝𝐱 𝟎 𝟎 Solution: Given order 𝑑𝑦𝑑𝑥 is in correct form. Given limits are y : 0→ √2𝑎𝑥 − 𝑥2 , x : 0→ 2𝑎 Equations are y = 0 , y = √2𝑎𝑥 − 𝑥2 , x = 0 , x = 2𝑎 y2 = 2ax − x2 x2 + y2 − 2ax = 0 is a circle with centre (a,0) and radius ‘a’. Replacement: Put x2 + y2 = r2 , 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑𝜃 Limits: r : 0 → 2𝑎𝑐𝑜𝑠𝜃 , 𝜃: 0 → 𝜋 2 2𝑎𝑐𝑜𝑠𝜃 𝑟2 × 𝑟𝑑𝑟𝑑𝜃 𝜋 ∫ 2𝑎 ∫ √2𝑎𝑥−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 = ∫2 ∫ 0 0 0 0 3 𝑟 𝑑𝑟𝑑𝜃 2𝑎𝑐𝑜𝑠 𝜃 𝜋 = ∫2 ∫ 0 0 4 ] 0 r4 2acosθ π = ∫2 [ 0 dθ 4 π 24a4cos4θ = ∫2( 0 − 0) dθ π 0 = 4a4 ∫2 cos4θdθ ENGG MATHS-II DOUBLE INTEGRATION
  • 6.
    = 4a4 ×3 × 1 × π 4 2 2 (∵ ∫ 4 π 2 0 cos θdθ = × × 3 1 π 4 2 2 ) = 3πa4 4 Example: 𝐱 𝐱𝟐+𝐲𝟐 𝐝𝐱𝐝𝐲 By changing into polar co-ordinates and evaluate∫ 𝟐 ∫ √𝟐𝐱−𝐱𝟐 𝟎 𝟎 Solution: Given order 𝑑𝑥𝑑𝑦 is in incorrect form. X X2+y2 dydx The correct form is 𝑑𝑦𝑑𝑥 ⇒ ∫ 2 ∫ √2X−X2 0 0 Given limits are y : 0→ √2𝑥 − 𝑥2 , x : 0→ 2 Equations are y = 0 , y = √2𝑥 − 𝑥2 , x = 0 , x = 2 y2 = 2x − x2 x2 + y2 − 2x = 0 is a circle with centre (1,0) and radius ‘1’. Replacement: Put x = rcos𝜃, 𝑥2 + 𝑦2 = 𝑟2 , dxdy = rdrd𝜃 Limits: r : 0 → 2𝑐𝑜𝑠𝜃 , 𝜃: 0 → 𝜋 2 𝑥 +𝑦 2 2 𝑑𝑦𝑑𝑥 = ∫ ∫ 2 √2𝑥−𝑥2 𝑥 ∫0∫0 𝑟2 2𝑐𝑜𝑠𝜃 𝑟𝑐𝑜𝑠𝜃 × 𝑟𝑑𝑟𝑑𝜃 𝜋 2 0 0 0 𝜋 0 =∫2[𝑟𝑐𝑜𝑠𝜃]2𝑐𝑜𝑠𝜃 𝑑𝜃 𝜋 0 = ∫2(2𝑐𝑜𝑠2𝜃 − 0)𝑑𝜃 ENGG MATHS-II DOUBLE INTEGRATION
  • 7.
    𝜋 0 = 2 ∫2𝑐𝑜𝑠2𝜃 𝑑𝜃 = 2 × 1 × 𝜋 2 2 (∵ ∫ 2 𝜋 2 0 1 𝜋 2 2 𝑐𝑜𝑠 𝜃𝑑𝜃 = × ) = 𝜋 2 Example: By changing into polar co-ordinates and evaluate∫ 𝐚 ∫ √𝐚𝟐−𝐱𝟐 √𝐱𝟐 + 𝐲𝟐𝐝𝐲𝐝𝐱 𝟎 𝟎 Solution: Given order 𝑑𝑥𝑑𝑦 is in correct form. Given limits are y : 0→ √𝑎2 − 𝑥2 , x : 0→ 𝑎 Equations are y = 0 , y = √𝑎2 − 𝑥2 , x = 0 , x = 𝑎 y2 = a2 − x2 x2 + y2 = a2 is a circle with centre (0,0) and radius ‘a’. Replacement: Put x2 + y2 = r2 ⇒ r = √x2 + y2 , 𝑑𝑦𝑑𝑥 = 𝑟𝑑𝑟𝑑𝜃 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 a r × rdrdθ 2 π ∫ a ∫ √a2−X2 √x2 + y2dydx = ∫2 ∫ 0 0 0 0 a r2drdθ π = ∫2 ∫ 0 0 3 π r3 a = ∫2 [ 0 ] dθ 0 ENGG MATHS-II DOUBLE INTEGRATION
  • 8.
    a3 3 π = ∫2 ( 0 −0)dθ = a3 3 π ∫2 dθ 0 a3 3 0 π = (θ)2 3 2 = a3 ( π − 0) = πa3 6 Example: 𝐱𝟐+𝐲𝟐 Evaluate ∬ 𝐱𝟐𝐲𝟐 𝐝𝐱𝐝𝐲 over the annular region between the circles x2 + y2 = a2 and x2 + y2 = b2 (b>a) by transforming into polar co-ordinates. Solution: 𝑥2+𝑦2 𝑟2 Replacement: Put x2 = r2cos2θ,y2 = r2sin2θ x2 + y2 = r2, 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑𝜃 Given the region is between the circles x2 + y2 = a2 and x2 + y2 = b2 Limits: r : 𝑎 → 𝑏 , 𝜃: 0 → 2𝜋 ∴ ∬ 𝑥2𝑦2 𝑑𝑥𝑑𝑦 = ∫ 2𝜋 ∫ 𝑏 𝑟2𝑐𝑜𝑠2𝜃×𝑟2𝑠𝑖𝑛2𝜃 × 𝑟𝑑𝑟𝑑𝜃 0 𝑎 ENGG MATHS-II DOUBLE INTEGRATION
  • 9.
    r2 0 a ENGG MATHS-II =∫ 2π ∫ b r5cos2θ×sin2θ × drdθ = ∫ 2π ∫ b r3cos2θ × sin2θ drdθ 0 a 4 a r4 b 2π 2 2 = ∫0 [ ] cos θ × sin θ dθ 4 0 = 1 ∫ 2π (b4 − a4) cos2θ × sin2θ dθ 4 = (b4−a4) ∫ 2π cos2θ × sin2θ dθ 0 = 4 4 (b −a ) 4 × ∫ π 2 4 0 2π 0 cos2θ × sin2θ dθ (∵ ∫ = 4∫ ) π 2 0 π = ( b4 − a4) × ∫2 cos2θ × sin2θ dθ 0 4 2 2 = ( b4 − a4) × 1 × 1 × π (∵ ∫ 2 2 π 2 0 1 1 π 4 2 2 cos θsin θdθ = × × ) = π(b4−a4) 16 Example: 𝟎 𝟎 Evaluate ∫ 𝐚 ∫ √𝐚𝟐−𝐱𝟐 √𝐚𝟐 − 𝐱𝟐 − 𝐲𝟐 𝐝𝐲𝐝𝐱 by transforming into polar co-ordinates. Solution: Given order 𝑑𝑦𝑑𝑥 is in correct form. Given limits are y : 0→ √𝑎2 − 𝑥2 , x : 0→ 𝑎 Equations are y = 0 , y = √𝑎2 − 𝑥2 , x = 0 , x = 𝑎 𝑦2 = 𝑎2 − 𝑥2 x2 + y2 = a2 is a circle with centre (0,0) and radius ‘a’. DOUBLE INTEGRATION
  • 10.
    ENGG MATHS-II Replacement: Put a2− x2 − y2 = a2 − (x2 + y2) = a2 − r2 , 𝑑𝑦𝑑𝑥 = 𝑟𝑑𝑟𝑑𝜃 ∴ √a2 − x2 − y2 = √a2 − r2 Limits: r : 0 → 𝑎 , 𝜃: 0 → 𝜋 2 𝑎 √𝑎2 − 𝑟2 𝑟𝑑𝑟𝑑𝜃 𝜋 ∫ 𝑎 ∫ √𝑎2−𝑥2 √𝑎2 − 𝑥2 − 𝑦2 𝑑𝑦𝑑𝑥 = ∫2 ∫ 0 0 0 0 = 2 2 − 𝑟 𝑟𝑑𝑟)𝑑𝜃 𝑎 𝜋 ∫2(∫ √𝑎 0 0 Substitution: Put a2 − r2 = t −2rdr = dt if r = 0 ⇒ t = a2 if r = a ⇒ t = 0 rdr = − dt 2 ∴ t ∶ a2 → 0 2 2 a π ∫2(∫ √a − r 0 0 −dt 2 0 π rdr)dθ = ∫2[∫ 2 √t( )]dθ 0 a 2 −1 0 π = ∫2[∫ 2 √tdt]dθ 0 a = −1 2 1 0 π ∫2[∫ 2 t2dt]dθ 0 a = −1 2 ∫ t 3 2 3 2 a2 0 π 2 0 [ ] dθ = − × 1 2 2 3 π 3 0 0 a2 ∫2 [t2] dθ 1 3 3 2 π = − ∫2(0 − (a )2)dθ = − 1 3 ∫ 0 π 3 − (a ) dθ 2 0 = a3 3 π ∫2 dθ 0 a3 3 0 π = (θ)2 DOUBLE INTEGRATION
  • 11.
    3 2 = a3 ( π − 0) = πa3 6 Exercise: Evaluatethe following by changing into polar co-ordinates. 1. ∫ 2𝑎 ∫ √2𝑥−𝑥2 𝑑𝑦𝑑𝑥 0 0 Ans: 𝜋𝑎2 2 2. ∫ 𝑎 ∫ √𝑎2−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 0 0 Ans: 𝜋𝑎4 8 𝑥2 3. ∫ 1 ∫ 2−𝑥 𝑥𝑦 𝑑𝑥𝑑𝑦 0 Ans: 3 8 𝑥 𝑑𝑥𝑑𝑦 𝑦 𝑥2+𝑦2 4. ∫ 𝑎 ∫ 𝑎 0 Ans: 𝜋𝑎 4 𝑥2+𝑦2 𝑑𝑥𝑑𝑦 5. ∫ 2𝑎 ∫ √2𝑎𝑥−𝑥2 𝑥 0 0 Ans: 𝜋𝑎 2 6. ∫ 𝑎 ∫ √𝑎2−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 −𝑎 0 Ans: 𝜋𝑎4 4 7. ∫ 𝑎 ∫ √𝑎2−𝑥2 (𝑥2𝑦 + 𝑦3)𝑑𝑥𝑑𝑦 0 0 Ans: 𝑎5 5 8. ∫ 1 ∫ √2𝑥−𝑥2 (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥 0 0 8 Ans: 3𝜋 − 1 X2+y2 9. ∬ X2y2 dxdy over the annular region between the circles x2 + y2 = 16 and x2 + y2 = 4 Ans:15π 10.∬ 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 – ordinates. To convert from Cartesian to cylindrical polar coordinates system we have the following transformation. 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑦 = 𝑟 𝑠𝑖𝑛𝜃 𝑧 = 𝑧 J = ∂(X ,y,z) = r ∂(r ,θ,z) Hence the integral becomes ENGG MATHS-II DOUBLE INTEGRATION
  • 12.
    ENGG MATHS-II ∭ f(x, y, z) dzdydx = ∭ f(r , θ, z) dzdrdθ Example: Find the volume of a solid bounded by the spherical surface 𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 = 𝟒𝐚𝟐 and the cylinder 𝐱𝟐 + 𝐲𝟐 − 𝟐𝐚𝐲 = 𝟎. Solution: Cylindrical co – ordinates x = r cos θ y = r sin θ z = z The equation of the sphere x2 + y2 + z2 = 4a2 r2cos2 θ + r2 sin2 θ + z2 = 4a2 r2 + z2 = 4a2 And the cylinder x2 + y2 − 2ay = 0 x2 + y2 = 2ay r2cos2 θ + r2 sin2 θ = 2a r sin θ r2 = 2arsin θ r = 2asin θ Hence, the required volume, DOUBLE INTEGRATION
  • 13.
    ∫ √4a2− r2 r dzdr dθ 0 r√4a2 − r2 dr dθ Volume = ∫ ∫ ∫ dx dy dz = ∫ ∫ ∫ r dθ dr dz = 4 ∫ π⁄2 ∫ 2a sin θ 0 0 = 4 ∫ π⁄2 ∫ 2a sin θ 0 0 3 0 = 4 ∫ π⁄2 [− 1 (4a2 − r2)3⁄2] 2a sin θ dθ 0 3 0 = 4 ∫ π⁄2 [−(4a2 − 4a2 sin2 θ)3⁄2 + 8a3] dθ = 4 3 ∫ π⁄2 (−8a3 cos3 θ + 8a3) dθ 0 3 = 4 8a3 ∫ π⁄2 (1 − cos3 θ) dθ 0 3 2 3 = 32 a3 [ π − 2 ] cubic units Example: Find the volume of the portion of the cylinder 𝐱𝟐 + 𝐲𝟐 = 𝟏 intercepted between the plane 𝐱 = 𝟎 and the paraboloid 𝐱𝟐 + 𝐲𝟐 = 𝟒 − 𝐳. Solution: Cylindrical co – ordinates 𝑥 = 𝑟 𝑐𝑜𝑠 𝜃 𝑦 = 𝑟 𝑠𝑖𝑛 𝜃 𝑧 = 𝑧 Given x2 + y2 = 1 r2cos2 θ + r2 sin2 θ = 1 r2 = 1 r = ±1 Given x2 + y2 = 4 − z r2cos2 θ + r2 sin2 θ = 4 − z r2 = 4 − z z = 4 − r2 ENGG MATHS-II DOUBLE INTEGRATION
  • 14.
    ENGG MATHS-II Hence therequired volume Volume = ∫ ∫ ∫ r dz dr dθ ∫ 4− r2 r dz dr dθ 4− r 0 2 dr dθ r (4 − r2) dr dθ (4r − r3) dr dθ = ∫ 2π ∫ 1 0 0 0 = ∫ 2π ∫ 1 r [z] 0 0 = ∫ 2π ∫ 1 0 0 = ∫ 2π ∫ 1 0 0 2 r4 − 4 ] dθ 2π [4r2 = ∫0 4 [(2 − 1 ) − (0 − 0)] dθ = ∫ 2π 0 4 7 dθ = ∫ 2π 0 4 0 = 7 [θ] 2π = 7 [2π − 0] 4 2 = 7 π cubic.units Example: Find the volume bounded by the paraboloid 𝐱𝟐 + 𝐲𝟐 = 𝐚𝐳, and the cylinder 𝐱𝟐 + 𝐲𝟐 = 𝟐𝐚𝐲 and the plane 𝐳 = 𝟎 Solution: Cylindrical co – ordinates x = r cos θ y = r sin θ z = z The equation of the sphere x2 + y2+= az r2cos2 θ + r2 sin2 θ = az r2 = az And the cylinder x2 + y2 = 2ay r2cos2 θ + r2 sin2 θ = 2a r sin θ r2 = 2arsin θ DOUBLE INTEGRATION
  • 15.
    r = 2asinθ Hence, the required volume, Volume = ∫ ∫ ∫ dx dy dz = ∫ ∫ ∫ r dθ dr dz r2 0 π 2a sin θ =∫ 0 ∫ 0 ∫ a r dz dr dθ 0 r2 [z]a rdrdθ ∫ 2a sin θ =∫ π 0 0 r3 [ a ] drdθ π 2a sin θ =∫0 ∫0 = 1 ∫ a 4 0 r4 2a sin θ π 0 [ ] dθ a 4 0 = 1 ∫ π 16a4sin4θ dθ π 2 = 4a3 × 2 ∫ ⁄ sin4θdθ 0 = 4a3 × 2 3 1 π = 3πa3 4 2 2 2 Change of variables from Cartesian Co – ordinates to spherical Polar Co – ordinates To convert from Cartesian to spherical polar co-ordinates system we have the following transformation 𝑥 = 𝑟𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜑 y = rsinθsinφ z = rcosθ J = ∂(r ,θ,φ ) ∂(X ,y,z) = r2sinθ Hence the integral becomes ∭ f(x , y, z) dzdydx = ∭ f(r , θ, z)r2sinθdrdθdφ Example: Evaluate ∫ ∫ ∫ 𝟏 √𝟏− 𝐱𝟐− 𝐲𝟐− 𝐳𝟐 𝐝𝐱 𝐝𝐲 𝐝𝐳 over the region bounded by the sphere 𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 = 𝟏. Solution: Let us transform this integral in spherical polar co – ordinates by taking ENGG MATHS-II DOUBLE INTEGRATION
  • 16.
    x = rsin θ cos ϕ y = r sin θ sin ϕ z = r cos θ dx dy dz = (r2 sin θ) dr dθ dϕ Hence ϕ varies from 0 to 2π ϕ varies from 0 to π ϕ varies from 0 to 1 r2 sin θ dr dθ d ϕ 0 √1− r2 = ∫ 2π ∫ π ∫ 1 1 0 0 0 0 r2 = [∫ 2π dϕ ] [∫ π sin θ dθ] [∫ 1 0 √1− r2 dr] 0 0 r2 dr = [ϕ] 2π [– cosθ] π ∫ 1 0 √1 − r2 r2 dr = (2π − 0) ( 1 + 1) ∫ 1 0 √1 − r2 r2 = 4π ∫ 1 dr 0 √1 − r2 Put r = sint ; dr = cost dt r = 0 ⇒ t = 0 r = 1 ⇒ t = π 2 sin2 t √1 − sin2t cost dt = 4π ∫ π⁄2 0 √ cos2t = 4π ∫ π⁄2 sin2 t cost dt 0 cost = 4π ∫ π⁄2 sin2 t cos t dt 0 = 4π ∫ π⁄2 sin2 t dt 0 2 2 = 4π 1 π = π2 Example: ∫√𝐱𝟐+𝐲𝟐 𝐝𝐳 𝐝𝐲 𝐝𝐱 √𝐱𝟐+ 𝐲𝟐+ 𝐳𝟐 𝟏 √𝟏− 𝐱𝟐 𝟏 Evaluate ∫𝟎 ∫𝟎 Solution: ENGG MATHS-II DOUBLE INTEGRATION
  • 17.
    Given 𝑥 variesfrom 0 to 1 y varies from 0 to √1 − x2 z varies from √x2 + y2 to 1 Let us transform this integral into spherical polar co – ordinates by using x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ dx dy dz = (r2 sin θ) dr dθ dϕ Let z = √x2 + y2 ⇒ z2 = x2 + y2 ⇒ r2 cos2 θ = r2 sin2θ cos2ϕ + r2 sin2θ sin2ϕ ⇒ cos2 θ = sin2 θ [∵ cos2ϕ + sin2ϕ = 1 ] ⇒ θ = π 4 Let 𝑧 = 1 ⇒ 𝑟 𝑐𝑜𝑠 𝜃 = 1 ⇒ 𝑟 = 1 𝑐𝑜𝑠 𝜃 ⇒ 𝑟 = 𝑠𝑒𝑐 𝜃 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 θ θ = 0 to θ = π 4 Limits of r ∶ Limits of θ ∶ Limits of ϕ ∶ ϕ = 0 to ϕ = π 2 r = ∫ π⁄2 ∫ π⁄4 ∫ sec θ 1 r2 sin θ dr dθ dϕ 0 0 0 = ∫ π⁄2 ∫ π⁄4 ∫ sec θ r sin θ dr dθ dϕ 0 0 0 r2 sec θ π⁄2 π⁄4 = ∫0 ∫0 [sin θ 2 ] 0 dθ dϕ 2 π⁄2 π⁄4 sec2θ sin θ−0 = ∫0 ∫0 [ ] dθ dϕ 2 0 0 = ∫ π⁄2 ∫ π⁄4 1 sec θ tan θ dθ dϕ 2 0 0 = [1 ∫ π⁄2 dϕ ] [∫ π⁄4 sec θ tan θ dθ ] ENGG MATHS-II DOUBLE INTEGRATION
  • 18.
    2 0 0 = 1 [θ ] π⁄2 [secθ] π⁄4 2 2 = 1 [ π − 0] [√2 − 1] 4 = π (√2 − 1) Example: Evaluate ∫ ∫ ∫ (𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 )𝐝𝐱 𝐝𝐲 𝐝𝐳 taken over the region bounded by the volume enclosed by the sphere 𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 = 𝟏. Solution: Let us convert the given integral into spherical polar co – ordinates. x = r sin θ cos ϕ ⇒ x2 = r2 sin2 θ cos2 ϕ y = r sin θ sin ϕ ⇒ y2 = r2 sin2 θ sin2 ϕ z = r cos θ ⇒ z2 = r2 cos2 θ dx dy dz = (r2 sin θ) dr dθ dϕ ∫ ∫ ∫ (x2 + y2 + z2 )dx dy dz = ∫ π ∫ 2π ∫ 1 r2 (r2 sin θ dθ dϕ dr) 0 0 0 Limits of r ∶ Limits of θ ∶ Limits of ϕ ∶ r = 0 to r = 1 θ = 0 to θ = π ϕ = 0 to ϕ = 2π ∫ ∫ ∫ (x2 + y2 + z2 )dx dy dz = ∫ π ∫ 2π ∫ 1 r2 (r2 sin θ dθ dϕ dr) 0 0 0 = [∫ 1 r4 dr] [∫ π sin θ dθ] [∫ 2π dϕ] 0 0 0 r5 1 0 π = [ 5 ] 0 [– cos θ] [ϕ]2π 0 5 = ( 1 − 0) (1 + 1) (2π − 0) = ( 1 ) (2) (2π) = 4π 5 5 ENGG MATHS-II DOUBLE INTEGRATION