Answer:
S = 2π/9 [³/₂√10 − ½ ln(-3 + √10)]
S ≈ 3.946
Step-by-step explanation:
For a curve rotated about the x-axis, the surface area is:
S = ∫ₐᵇ 2πy ds,
where ds = √(1 + (dy/dx)²) dx.
y = e⁻³ˣ
dy/dx = -3e⁻³ˣ
ds = √(1 + (-3e⁻³ˣ)²) dx
S = ∫₀°° 2πe⁻³ˣ √(1 + (-3e⁻³ˣ)²) dx
If u = -3e⁻³ˣ, then du = 9e⁻³ˣ dx, or du/9 = e⁻³ˣ dx.
When x = 0, u = -3. When x = ∞, u = 0.
S = ∫₋₃⁰ 2π √(1 + u²) (du/9)
S = 2π/9 ∫₋₃⁰ √(1 + u²) du
S = 2π/9 [ ½ u √(1 + u²) + ½ ln|u + √(1 + u²)| ] |₋₃⁰
S = 2π/9 {[0] − [ -³/₂√10 + ½ ln(-3 + √10) ]}
S = 2π/9 [³/₂√10 − ½ ln(-3 + √10)]
S ≈ 3.946
The area of a surface is the amount of space it occupies.
The area of the resulting surface is [tex]3.947[/tex] square units
The infinite curve is given as:
[tex]y =e^{-3x},\ \ x \ge 0[/tex]
Integrate y, with respect to x
[tex]\frac{dy}{dx} = -3e^{-3x}[/tex]
The area of the curve about the x-axis is:
[tex]S = \int\limits^a_b {2\pi y} \, ds[/tex]
[tex]ds = \sqrt{(1 + (\frac{dy}{dx})^2)}\ dx[/tex]
Substitute [tex]\frac{dy}{dx} = -3e^{-3x}[/tex] in [tex]ds = \sqrt{(1 + (\frac{dy}{dx})^2)}\ dx[/tex]
[tex]ds = \sqrt{(1 + (-3e^{-3x})^2)}\ dx[/tex]
Let
[tex]u = -3e^{-3x}[/tex]
So:
[tex]\frac{du}{dx} = 9e^{-3x}[/tex]
Make [tex]e^{-3x}\ dx[/tex] the subject
[tex]e^{-3x}\ dx = \frac{du}9[/tex]
[tex]x \ge 0[/tex] means that, the value of x is: [tex][0,\infty][/tex]
When [tex]x = 0[/tex]
[tex]u = -3e^{-3x}[/tex]
[tex]u = -3 \times e^{-3 \times 0} = -3[/tex]
When [tex]x = \infty[/tex]
[tex]u = -3 \times e^{-3 \times \infty} = 0[/tex]
Recall that:
[tex]S = \int\limits^a_b {2\pi y} \, ds[/tex]
Substitute [tex]ds = \sqrt{(1 + (-3e^{-3x})^2)}\ dx[/tex] and [tex]y =e^{-3x}[/tex]
This gives
[tex]S = \int\limits^0_{-3} {2\pi (e^{-3x}) \sqrt{(1 + (-3e^{-3x})^2)}\ dx}[/tex]
Rewrite as:
[tex]S = \int\limits^0_{-3} {2\pi \sqrt{(1 + (-3e^{-3x})^2)}\ (e^{-3x})\ dx}[/tex]
Substitute [tex]u = -3e^{-3x}[/tex] and [tex]e^{-3x}\ dx = \frac{du}9[/tex]
[tex]S = \int\limits^0_{-3} {2\pi \sqrt{(1 + u^2)}\ \frac{du}9}[/tex]
This gives
[tex]S = \frac{2\pi}{9} \int\limits^0_{-3} {\sqrt{(1 + u^2)}\ du}[/tex]
Integrate with respect to u
[tex]S = \frac{2\pi}{9}[\frac 12 u\sqrt{(1 + u^2)} + \frac 12\ln|u + \sqrt{1 + u^2}|\ ]|\limits^0_{-3}[/tex]
Substitute 0 and -3 for u
[tex]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} ] )[/tex]
[tex]S = \frac{2\pi}{9}([0] - [\frac 12 \times (-3) \times \sqrt{(1 + (-3)^2)} + \frac 12\ln|-3 + \sqrt{1 + (-3)^2} ] )[/tex]
[tex]S = \frac{2\pi}{9}(- [\frac 12 \times (-3) \times \sqrt{(1 + (-3)^2)} + \frac 12\ln|-3 + \sqrt{1 + (-3)^2} ] )[/tex]
[tex]S = \frac{2\pi}{9}(- [\frac{-3}2 \times \sqrt{(1 + (-3)^2)} + \frac 12\ln|-3 + \sqrt{1 + (-3)^2} ] )[/tex]
[tex]S = \frac{2\pi}{9}( [\frac{3}2 \times \sqrt{10} - \frac 12\ln(-3 + \sqrt{10}\ )] )[/tex]
[tex]S = \frac{2\pi}{9}( [4.743 + 0.909] )[/tex]
[tex]S = \frac{2\pi}{9}( 5.652 )[/tex]
[tex]S = 3.947[/tex]
Hence, the area of the resulting surface is [tex]3.947[/tex] square units
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