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Let [tex]C[/tex] be the curve. The surface area upon revolving the bounded region is given by

[tex]\displaystyle2\pi\int_C y\,\mathrm dS=2\pi\int_0^\infty y\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=2\pi\int_0^\infty e^{-9x}\sqrt{1+(-9e^{-9x})^2}\,\mathrm dx[/tex]
[tex]=2\pi\displaystyle\int_0^\infty e^{-9x}\sqrt{1+81e^{-18x}}\,\mathrm dx[/tex]

Let [tex]u=e^{-9x}[/tex] so that [tex]\mathrm du=-9e^{-9x}\,\mathrm dx[/tex]. Then the integral becomes

[tex]-\dfrac{2\pi}9\int_1^0\sqrt{1+81u^2}\,\mathrm du=\dfrac{2\pi}9\int_0^1\sqrt{1+81u^2}\,\mathrm du[/tex]

Now take [tex]u=\dfrac19\tan v[/tex], so that [tex]\mathrm du=\dfrac19\sec^2v\,\mathrm dv[/tex] and we have

[tex]\displaystyle\dfrac{2\pi}{81}\int_0^{\arctan9}\sqrt{1+81\left(\frac19\tan v\right)^2}\sec^2v\,\mathrm dv[/tex]
[tex]=\displaystyle\dfrac{2\pi}{81}\int_0^{\arctan9}\sec^3v\,\mathrm dv[/tex]
[tex]=\dfrac{\pi\sqrt{82}}9+\dfrac\pi{81}\ln(9+\sqrt{82})[/tex]

The infinite curve [tex]\rm y = e^{-9x}[/tex], x ≥ 0, is rotated about the x-axis then the area of the resulting surface is,

[tex]\dfrac{\pi \sqrt{82}}{9} + \dfrac{\pi }{81} ln(9 + \sqrt{81})[/tex]

What is the area of revolution?

The surface area that is generated by the revolution of the curve and bounded by the curve is called the area of revolution.

The infinite curve is rotated about the x-axis.

[tex]\rm y = e^{-9x}, x \geq 0,[/tex]

We know that the equation of the surface area

[tex]\rm Surface\ area = 2 \pi \int_C ydS \\\\ Surface\ area = 2 \pi \int_0^{\infty} y \sqrt{1 + (\dfrac{dy}{dx})^2}\ \ dx =\\\\Surface\ area = 2 \pi \int_0^{\infty} e^{-9x} \sqrt{1+ (-9e^{-9x})^2} \ dx\\\\Surface \ area = 2\pi \int_0^{\infty} e^{-9x} \sqrt{1+81e^{-18x}} \ dx[/tex]

[tex]\rm Let, \ u =e^{-9x}\\\\Then \ du = -9e^{-9x} dx[/tex]

Put these values in the equation. Then integrate it

[tex]\rm Surface \ area = 2\pi \int_0^{\infty} e^{-9x} \sqrt{1+81e^{-18x}} \ dx \\\\Surface \ area = - \dfrac{2\pi}{9} \int _1^0 \sqrt{1 + 81 u^2} \ du \\\\Surface \ area = \dfrac{2\pi}{9} \int _0^1 \sqrt{1 + 81 u^2} \ du[/tex]

Now take 9u = tan v, so that 9du = sec² v dv, then we have

[tex]\rm Surface \ area = \dfrac{2\pi}{81} \int _0^{arctan\ 9 } \sqrt{1 + 81 (\dfrac{1}{9} tan \ v )^2} sec^2 v \ dv \\\\Surface \ area = \dfrac{2\pi}{81} \int _0^{arctan\ 9 } sec^3 v dv\\\\Surface \ area = \dfrac{\pi \sqrt{82}}{9} + \dfrac{\pi }{81} ln(9 + \sqrt{81})[/tex]

Thus, the area of the resulting surface is,

[tex]\rm Surface \ area = \dfrac{\pi \sqrt{82}}{9} + \dfrac{\pi }{81} ln(9 + \sqrt{81})[/tex]

More about the area of revolution link is given below.

https://brainly.com/question/26166671

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