Mathematics, 15.07.2020 02:01 mrfishyyyy

If the infinite curve y = e^−3x, x ≥ 0, is rotated about the x-axis, find the area of the resulting surface.

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Answer from: coolstacie

The area of a surface is the amount of space it occupies.

The area of the resulting surface is 3.947 square units

The infinite curve is given as:

$y =e^{-3x},\ \ x \ge 0$

Integrate y, with respect to x

$\frac{dy}{dx} = -3e^{-3x}$

The area of the curve about the x-axis is:

$S = \int\limits^a_b {2\pi y} \, ds$

$ds = \sqrt{(1 + (\frac{dy}{dx})^2)}\ dx$

Substitute $\frac{dy}{dx} = -3e^{-3x}$ in $ds = \sqrt{(1 + (\frac{dy}{dx})^2)}\ dx$

$ds = \sqrt{(1 + (-3e^{-3x})^2)}\ dx$

Let

$u = -3e^{-3x}$

So:

$\frac{du}{dx} = 9e^{-3x}$

Make $e^{-3x}\ dx$ the subject

$e^{-3x}\ dx = \frac{du}9$

$x \ge 0$ means that, the value of x is: $[0,\infty]$

When x = 0

$u = -3e^{-3x}$

$u = -3 \times e^{-3 \times 0} = -3$

When $x = \infty$

$u = -3 \times e^{-3 \times \infty} = 0$

Recall that:

$S = \int\limits^a_b {2\pi y} \, ds$

Substitute $ds = \sqrt{(1 + (-3e^{-3x})^2)}\ dx$ and $y =e^{-3x}$

This gives

$S = \int\limits^0_{-3} {2\pi (e^{-3x}) \sqrt{(1 + (-3e^{-3x})^2)}\ dx}$

Rewrite as:

$S = \int\limits^0_{-3} {2\pi \sqrt{(1 + (-3e^{-3x})^2)}\ (e^{-3x})\ dx}$

Substitute $u = -3e^{-3x}$ and $e^{-3x}\ dx = \frac{du}9$

$S = \int\limits^0_{-3} {2\pi \sqrt{(1 + u^2)}\ \frac{du}9}$

This gives

$S = \frac{2\pi}{9} \int\limits^0_{-3} {\sqrt{(1 + u^2)}\ du}$

Integrate with respect to u

$S = \frac{2\pi}{9}[\frac 12 u\sqrt{(1 + u^2)} + \frac 12\ln|u + \sqrt{1 + u^2}|\ ]|\limits^0_{-3}$

Substitute 0 and -3 for u

$S = \frac{2\pi}{9}([\frac 12\times 0 \times \sqrt{(1 + 0^2)} + \frac 12\ln|0 + \sqrt{1 + 0^2} ] - [\frac 12 \times (-3) \times \sqrt{(1 + (-3)^2)} + \frac 12\ln|-3 + \sqrt{1 + (-3)^2} ] )$

$S = \frac{2\pi}{9}([0] - [\frac 12 \times (-3) \times \sqrt{(1 + (-3)^2)} + \frac 12\ln|-3 + \sqrt{1 + (-3)^2} ] )$

$S = \frac{2\pi}{9}(- [\frac 12 \times (-3) \times \sqrt{(1 + (-3)^2)} + \frac 12\ln|-3 + \sqrt{1 + (-3)^2} ] )$

$S = \frac{2\pi}{9}(- [\frac{-3}2 \times \sqrt{(1 + (-3)^2)} + \frac 12\ln|-3 + \sqrt{1 + (-3)^2} ] )$

$S = \frac{2\pi}{9}( [\frac{3}2 \times \sqrt{10} - \frac 12\ln(-3 + \sqrt{10}\ )] )$

$S = \frac{2\pi}{9}( [4.743 + 0.909] )$

$S = \frac{2\pi}{9}( 5.652 )$

S = 3.947

Hence, the area of the resulting surface is 3.947 square units

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If the infinite curve y = e^−3x, x ≥ 0, is rotated about the x-axis, find the area of the resulting...

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