Subject:
SATAuthor:
madisynballCreated:
1 year agoWe have
x(t) = t³ ⇒ dx/dt = 3t²
y(t) = t ⇒ dy/dt = 1
so that the line element is
ds = √((dx/dt)² + (dy/dt)²) dt = √(9t⁴ + 1) dt
and the line integral is
\displaystyle \int_C y^3 \, ds = \int_0^5 t^3 \sqrt{9t^4 + 1} \, dt
Substitute u = 9t⁴ + 1 and du = 36t³ dt :
\displaystyle \int_C y^3 \, ds = \frac1{36} \int_1^{5626} \sqrt u \, du = \frac1{36} \cdot \frac23 \left((5626^{\frac32} - 1^{\frac32}\right) = \boxed{\frac{5626^{\frac32} - 1}{54}}
Author:
amigavuii
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