Question

Evaluate the line integral, where C is the given curve. \int_{C}xy\ ds C:x=t^{2}, y=2t, 0\leq t\leq4

Integrals
ANSWERED
asked 2021-05-09
Evaluate the line integral, where C is the given curve.
\(\int_{C}xy\ ds\)
\(C:x=t^{2},\)
\(y=2t,\)
\(0\leq t\leq4\)

Answers (1)

2021-05-10

Step 1
Consider
\(x=t^{2}\)
\(\frac{dx}{dt}=2t\)
\(y=2t\)
\(\frac{dy}{dt}=2\)
Consider:
\(ds=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\cdot dt\)
\(=\sqrt{(2t)^{2}+(2)^{2}}\cdot dt\)
\(=\sqrt{4t^{2}+4}\cdot dt\)
\(=\sqrt{4(t^{2}+1)}\cdot dt\)
\(ds=2\sqrt{(t^{2}+1)}\cdot dt\)
Hence the line integral is,
\(\int_{C}xy\ ds=\int_{t=0}^{4}(t)^{2}(2t)(2\sqrt{t^{2}+1}\ dt)\)
\(=4\int_{t=0}^{4}t^{3}\sqrt{t^{2}+1}\ dt\)
1) \(=4\int_{t=0}^{4}t^{2}\sqrt{t^{2}+1}t\cdot dt\)
Step 2
Let
\(t^{2}+1=u^{2}\)
\(2t\ dt=2u\ du\)
\(t\ dt=u\ du\)
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
\(\int_{C}xy\ ds=4\int_{t=0}^{4}t^{2}\sqrt{t^{2}+1}t\cdot dt\)
\(=4\int_{u=1}^{\sqrt{17}}(\sqrt{u^{2}})\times u\ du\) \(=4\int_{u=1}^{\sqrt{17}}(u^{2}-1)u\times u\ du\)
\(=4\int_{u=1}^{\sqrt{17}}(u^{2}-1)u^{2}\ du\)
\(=4\int_{u=1}^{\sqrt{17}}u^{4}-u^{2}\ du\)
\(=4\left[\frac{u^{5}}{5}-\frac{u^{3}}{3}\right]_{1}^{\sqrt{17}}\)
\(=4\left[\left(\frac{(\sqrt{17})^{5}}{5}-\frac{(\sqrt{17})^{3}}{3}\right)-\left(\frac{1}{5}-\frac{1}{3}\right)\right]\)
\(=4\left(\frac{782\sqrt{17}}{15}-\left(-\frac{2}{15}\right)\right)\)
\(\int_{C}xy\ ds=860.33\)

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