Resolution of Exam 3 - Analytical Geometry and Calculus 3 | MATH 2400, Exams of Analytical Geometry and Calculus

Download Resolution of Exam 3 - Analytical Geometry and Calculus 3 | MATH 2400 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Exam 3 Solutions 1. Evaluate the following integrals: (a) ∫ 4 −2 ∫ 3 0 ∫ 2 1 xz2 dy dz dx = (∫ 4 −2 x dx )(∫ 3 0 z2 dz )(∫ 2 1 dy ) = [ 1 2 x2 ]4 −2 · [ 1 3 ]3 0 · [ y ]2 1 = (8− 2) · (9− 0) · (2− 1) = 6 · 9 · 1 = 54. (b) ∫ 16 0 ∫ 4 √ y ex 3 dx dy =∫ 4 0 ∫ x2 0 ex 3 dy dx = ∫ 4 0 x2ex 3 dx = [ 1 3 ex 3 ]4 0 = e64 − 1 3 . (c) ∫ 1√ 2 0 ∫ √4−x2 √ 1−x2 √ x2 + y2 dy dx+ ∫ √2 1√ 2 ∫ √4−x2 x √ x2 + y2 dy dx = ∫ π 2 π 4 ∫ 2 1 √ r2 r dr dθ = π 4 [ 1 3 r3 ]2 1 = 7 12 π. (d) ∫ 2 −2 ∫ √4−x2 0 ∫ √4−x2−y2 0 √ x2 + y2 + z2 dz dy dx =∫ π 0 ∫ π 2 0 ∫ 2 0 √ ρ2ρ2 sinφ dρ dφ dθ = π [ − cosφ ]π 2 0 · [ 1 4 ρ4 ]2 0 = π · 1 · 4 = 4π. 2. Find the volume of the solid that is in the first octant, and bounded above by the plane x+ 2y + 3z = 6. ∫ 3 0 ∫ 6−2y 0 ( 2− 2 3 y − 1 3 x ) dx dy = ∫ 3 0 [ 2x− 2 3 xy − 1 6 x2 ]6−2y 0 dy = ∫ 3 0 ( 12− 4y − 4y + 4 3 y2 − 6 + 4y − 2 3 y2 ) dy = ∫ 3 0 ( 6− 4y + 2 3 y2 ) dy = [ 6y − 2y2 + 2 9 y3 ]3 0 = 18− 18 + 6 = 6. 3. Find the surface area of the surface defined by r(u, v) = 〈u cos v, u2, u sin v〉, 0 ≤ u ≤ √ 2, 0 ≤ v ≤ 2π. ru × rv = ∣∣∣∣∣∣ i j k cos v 2u sin v −u sin v 0 u cos v ∣∣∣∣∣∣ = 〈2u2 cos v,−(u cos2 v + u sin2 v), 2u2 sin v〉 = 〈2u2 cos v,−u, 2u2 sin v〉 ‖ru × rv‖ = √ 4u4 cos2 v + u2 + 4u4 sin2 u = u √ 4u2 + 1