Mathematics: Evaluate the line integral c y3 ds, c.x = t3, y = t, 0 â‰¤ t â‰¤ 3

Evaluate the line integral c y3 ds, c.x = t3, y =

Ответ:

missymiss4273

13.03.2022

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\int_{t=0}^{t=3}t^3\sqrt{1^2+(3t^2)^2}\,\mathrm dt=\int_0^3t^3\sqrt{1+9t^4}\,\mathrm dt$

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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Ответ:

taylordalton93

01.08.2021

Mathematics

x = 27

Step-by-step explanation:

3x -4y = 65

3x -4(4) = 65

3x - 16 = 65

+16 +16

16s cross out and 65 + 16 =81

3x = 81

3x/3 = 81/3

3s cross out

x = 27

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

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