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Please send the answers within 1 hour. Thanks.

Please Send The Answers Within 1 Hour Thanks Follow The Steps 1 (8.89 KiB) Viewed 20 times

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Evaluate the line integral fy(tanx + x sec² x) dx + xtanx dy + dz once around the circle x² + y² = 1, z = 0.

evaluate the line integral. 4. 6.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, = f'(² [(-2+ 3t)²(3 dt) + (3-3t)²(-3dt) + (3 - 3t)²(-3 dt)] √ 2² dx + y² dy + z² dz == 6. $x x²y dx + (x - y) dy once counterclockwise around the curve C bounding the region described by the curves x = 1 - y², y = x + 1 = [(-2+3t)²-2(3-3t)²] dt = 3 (-2+3t)³+(3-31)³ (3 — 31)³}" = 0 6. § x²y dx + (x −y) dy = √¸ x²y dx + (x − y) dy + √₂ x³y dx + (x − y) dy = [', ((1 − 1²³)¹²t(−2t dt) + (1 - 1² -- t) dl] + www + √° [t² (1 – t) (−dt) + (−t − 1 + t) (−dt)] = ²₁ (-21³ + 47¹ - 36²2 - + + 1) dt + t - (-2/7 415 + = 2² +1}²_₂² - t + S² (8³ – 1² + 1) dt - 3 (4-5 3 +t 0 -- 99 140 (-3,-2) C₂ix=-1₂ y=1-t, 0€143 -15. Gix=1-1², yut, -2€141