Question

Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 4.

Answer 1

We have

x = t³   ===>   dx/dt = 3t²

y = t⁴   ===>   dy/dt = 4t³

Then with the given parameteriztion, the line integral along C of x/y is

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^3}{t^4} \sqrt{(3t^2)^2 + (4t^3)^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac1t \sqrt{9t^4 + 16t^6} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{\sqrt{t^4}}t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^2}t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 32t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 \sqrt{9 + 16t^2} \, d\left(9+16t^2\right)$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \cdot \frac23 \left(9+16t^2\right)^{\frac32}\bigg|_1^4$

$\displaystyle \int_C \frac xy \, ds = \frac1{48} \left(265^{\frac32} - 25^{\frac32}\right) = \boxed{\frac{265\sqrt{265}-125}{48}}$

Answer 2

The line integral will be "$\frac{265\sqrt{265} - 125 }{48}$". To understand the calculation, check below.

Line integral

According to the question,

Curves, x = t³ or,

            $\frac{dx}{dt}$ = 3t²

             y = t⁴ or,

            $\frac{dx}{dt}$ = 4t³  

Now, the line integral along C will be:

→ $\int\limits_C {\frac{x}{y} } \, ds$ = $\int\limits^4_1 {\frac{t^3}{t^4} }$ √(3t²)² + (4t³)² dt

             = $\int\limits^4_1 {\frac{1}{t} }$ √9t⁴ + 16t⁶ dt

             = $\int\limits^4_1 {\frac{t^4}{t} }$ √9 + 16t² dt

             = $\int\limits^4_1 {\frac{t^2}{t} }$ √9 + 16t² dt

             = $\int\limits^4_1$ t√9 + 16t² dt

             = $\frac{1}{32}$ $\int\limits^4_1$ √9 + 16t² d (9 + 16t²)

             = $\frac{1}{32}\times \frac{2}{3}$ $(9+16t^2)^{\frac{3}{2} } |_1^4$

             = $\frac{1}{48}$ ($265^{\frac{3}{2}} - 25^{\frac{3}{2}$)

             = $\frac{265 \sqrt{265 } -125 }{48}$

Thus the above approach is correct.

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