Please send the answers within 1 hour. Thanks. Follow the Steps:
Posted: Mon Jul 11, 2022 11:02 am
Please send the answers within 1 hour. Thanks.
Follow the Steps:
[a] Find a parametric representation for the line segment from (0, 3) to (4, 6). Using the parametric equations in part [a] to evaluate the line integral Jo where C is the line segment in part [a]. x sin(y) ds
evaluate the line integral. 4. [.x²6 ² dx + y² dy + z² dz, where C is the curve x + y = 1, .x+ z = 1 from (-2, 3, 3) to (1, 0, 0) 4. With parametric equations C: x=-2+3t, y=3-3t, z=3-3t, 0≤t≤ 1, for the straight line, [[ 2³² dx + y² dy + z² dz= = f'(² == [(-2+ 3t)²(3 dt) + (3-3t)²(-3dt) + (3 - 3t)²(-3 dt)] 3{/=(- = 3 ′ (−2 + 3t)³² — 2(3 — 3t)³] dt = 3 { = - - (-2+3t)³ + (3 — 32)³} = 0 2 -15.
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evaluate the line integral. 4. [.x²6 ² dx + y² dy + z² dz, where C is the curve x + y = 1, .x+ z = 1 from (-2, 3, 3) to (1, 0, 0) 4. With parametric equations C: x=-2+3t, y=3-3t, z=3-3t, 0≤t≤ 1, for the straight line, [[ 2³² dx + y² dy + z² dz= = f'(² == [(-2+ 3t)²(3 dt) + (3-3t)²(-3dt) + (3 - 3t)²(-3 dt)] 3{/=(- = 3 ′ (−2 + 3t)³² — 2(3 — 3t)³] dt = 3 { = - - (-2+3t)³ + (3 — 32)³} = 0 2 -15.