Evaluate โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง , from ๐‘ง = 0 to ๐‘ง = 4+2๐‘– along the curve ๐ถ given by ๐‘ง = ๐‘กยฒ + ๐‘–๐‘ก .

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๐‘ธ๐’–๐’†๐’”๐’•๐’Š๐’๐’:-

Evaluate โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง , from ๐‘ง = 0 to ๐‘ง = 4+2๐‘– along the curve ๐ถ given by ๐‘ง = ๐‘กยฒ+๐‘–๐‘ก .

๐‘บ๐’๐’๐’–๐’•๐’Š๐’๐’:-

Given that the curve
๐‘ง = ๐‘กยฒ+๐‘–๐‘ก
โ‡’ ๐‘ฅ+๐‘–๐‘ฆ = ๐‘กยฒ+๐‘–๐‘ก
โ‡’ ๐‘ฅ = ๐‘กยฒ and ๐‘ฆ = ๐‘ก
โ‡’ ๐‘ฅ = ๐‘ฆยฒ this is the equation of parabola for the given parametric equation ๐‘ง = ๐‘กยฒ+๐‘–๐‘ก
โˆด ๐‘ง โ†’ 0 โ‡’ ๐‘ก โ†’ 0 and ๐‘ง โ†’ 4+2๐‘– โ‡’ ๐‘ก โ†’ 2
and we know that ๐‘ง = ๐‘กยฒ+๐‘–๐‘ก โˆด ๐‘‘๐‘ง=(2๐‘ก+๐‘–)๐‘‘๐‘ก
and ๐‘งฬ… = ๐‘กยฒ - ๐‘–๐‘ก
Now Putting in the given integral

โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง = โˆซ๊ฃขแคฒ(๐‘กยฒ+๐‘–๐‘ก)(2๐‘ก+๐‘–)๐‘‘๐‘ก
โ‡’โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง = โˆซ๊ฃขแคฒ(2๐‘กยณ+๐‘ก-๐‘–๐‘กยฒ)๐‘‘๐‘ก
โ‡’โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง = [2(๐‘กโด/4)+(๐‘กยฒ/2)]๊ฃขแคฒ - ๐‘– [(๐‘กยณ/3)]๊ฃขแคฒ
โ‡’โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง = [2(16/4)+(4/2) - 0 - 0] - ๐‘– [(8/3) - 0]
โ‡’โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง = [2(16/4)+(4/2) - 0 - 0] - ๐‘– [(8/3) - 0]
โ‡’โˆซ๐‘€พ ๐‘งฬ… ๐‘‘๐‘ง = 10 - (8/3)๐‘–
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Solved...!!!
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#complexanalysis #csirnet #complexintegration #riemannintegral
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