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2 years ago

Find the area between the graph please

Mathematics

1 answer:

vagabundo [1.1K]2 years ago

4 0

The area of the region shown in the coordinate plane is 2.7823 units square.

<h3>What is integration?</h3>

It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.

It is given that:

The graph of a function:

y = sinx

And a line segment passed through:

(0, 0) and (7π/6, -1/2)

First, find the equation of the line:

y = (-1/2)/(7π/6)[x]

y = (-3/7π)x

The area by definite integration:

$\rm Area = \int\limits^{7\pi/6}_0 {[sinx - (-3/7\pi)x]} \, dx$

$\rm Area = \int\limits^{7\pi/6}_0 {[sinx +3x/7\pi ]} \, dx$

After solving the above definite integral:

Area = 2.7823 units square

Thus, the area of the region shown in the coordinate plane is 2.7823 units square.

Learn more about integration here:

brainly.com/question/18125359

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gtnhenbr [62]

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3 years ago

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3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

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$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

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3 years ago