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Anuta_ua [19.1K]
3 years ago
6

Evaluate the line integral c y3 ds, c.x = t3, y = t, 0 ≤ t ≤ 3

Mathematics
1 answer:
Temka [501]3 years ago
7 0
\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\int_{t=0}^{t=3}t^3\sqrt{1^2+(3t^2)^2}\,\mathrm dt=\int_0^3t^3\sqrt{1+9t^4}\,\mathrm dt

Take u=1+9t^4 so that \mathrm du=36t^3\,\mathrm dt. Then

\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\frac1{36}\int_{u=1}^{u=730}\sqrt u\,\mathrm du=\frac{730^{3/2}-1}{54}
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