Evaluate the line integral, where C is the given curve. ∫Cxy ds C:x=t2, y=2t, 0≤t≤4

Expert Answer to "Evaluate the line integral, where C is the..."

waigaK

Answered question

2021-05-09

Evaluate the line integral, where C is the given curve.

\int_{C} x y d s

C : x = t^{2},

y = 2 t,

0 \leq t \leq 4

Answer & Explanation

mhalmantus

Skilled2021-05-10Added 105 answers

Step 1
Consider
$x = t^{2}$
$\frac{d x}{d t} = 2 t$
$y = 2 t$
$\frac{d y}{d t} = 2$
Consider:
$d s = \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} \cdot d t$
$= \sqrt{(2 t)^{2} + (2)^{2}} \cdot d t$
$= \sqrt{4 t^{2} + 4} \cdot d t$
$= \sqrt{4 (t^{2} + 1)} \cdot d t$
$d s = 2 \sqrt{(t^{2} + 1)} \cdot d t$
Hence the line integral is,
$\int_{C} x y d s = \int_{t = 0}^{4} (t)^{2} (2 t) (2 \sqrt{t^{2} + 1} d t)$
$= 4 \int_{t = 0}^{4} t^{3} \sqrt{t^{2} + 1} d t$
1) $= 4 \int_{t = 0}^{4} t^{2} \sqrt{t^{2} + 1} t \cdot d t$
Step 2
Let
$t^{2} + 1 = u^{2}$
$2 t d t = 2 u d u$
$t d t = u d u$
For
$t = 0$ then
$t^{2} + 1 = u^{2}$
$u^{2} = 1$
$u = 1$
For
$t = 4$ then
$t^{2} + 1 = u^{2}$
$u^{2} = 4^{2} + 1$
$u^{2} = 17$
$u = \sqrt{17}$
Substitute these values in (1) then
$\int_{C} x y d s = 4 \int_{t = 0}^{4} t^{2} \sqrt{t^{2} + 1} t \cdot d t$
$= 4 \int_{u = 1}^{\sqrt{17}} (\sqrt{u^{2}}) \times u d u$ $= 4 \int_{u = 1}^{\sqrt{17}} (u^{2} - 1) u \times u d u$
$= 4 \int_{u = 1}^{\sqrt{17}} (u^{2} - 1) u^{2} d u$
$= 4 \int_{u = 1}^{\sqrt{17}} u^{4} - u^{2} d u$
$= 4 {[\frac{u^{5}}{5} - \frac{u^{3}}{3}]}_{1}^{\sqrt{17}}$
$= 4 [(\frac{(\sqrt{17})^{5}}{5} - \frac{(\sqrt{17})^{3}}{3}) - (\frac{1}{5} - \frac{1}{3})]$
$= 4 (\frac{782 \sqrt{17}}{15} - (- \frac{2}{15}))$
$\int_{C} x y d s = 860.33$

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