Respuesta :

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Parameterize the contour by

[tex]C:\mathbf r(t)=(2\cos t,2\sin t)\implies \dfrac{\mathrm d\mathbf r}{\mathrm dt}=(-2\sin t,2\cos t)[/tex]
[tex]\implies\mathrm dS=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt=2[/tex]

where [tex]-\dfrac\pi2\le t\le\dfrac\pi2[/tex]. Then the line integral is given by

[tex]\displaystyle\int_C xy^2\,\mathrm dS=2\int_{-\pi/2}^{\pi/2}(2\cos t)(2\sin t)^2\,\mathrm dt[/tex]
[tex]=\displaystyle16\int_{-\pi/2}^{\pi/2}\cos t\sin^2t\,\mathrm dt=\dfrac{32}3[/tex]

The Right half of the circle as [tex]x^{2} + y^{2} = 4[/tex]  it is deduced that the Curve counterclockwise is [tex]\frac{32}{3}[/tex].

Given that,

Curve is [tex]c = xy^{2}ds[/tex]

Right half of the circle [tex]x^{2} + y^{2} = 4[/tex]

We have to evaluate the line integral.

According to the question,

Right half of the circle is counterclockwise incorrect

Therefore, [tex]x = 2cos\theta, y = 2sin\theta[/tex]

Then,

[tex]ds = \sqrt{x^{2} ++ y^{2} } \\ds = \sqrt{(2sin\theta)^{2}+ (2cos\theta)^{2}dt } \\ ds = \sqrt{2^{2}(sin^{2} \theta+ cos^{2} \theta )dt } \\ds = \sqrt{2^{2}.(1) dt} \\ds = 2dt[/tex]

The curve C,

[tex]c= \int\limits^\frac{\pi }{2} _\frac{-\pi }{2} {xy^{2} } \, ds\\c = \int\limits^\frac{\pi }{2} _\frac{-\pi }{2} {(2cost).(2sint)^{2} y^{2} } \, ds\\c = \frac{32}{3}[/tex]

Hence, The Right half of the circle as [tex]x^{2} + y^{2} = 4[/tex]  it is deduced that the Curve counterclockwise is [tex]\frac{32}{3}[/tex].

For more information about Integration click the link given below.

https://brainly.com/question/22008756

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