compute the surface area of revolution about the x-axis over the interval [0,1] for y=e^-3x

To compute the surface area of revolution about the x-axis for the function \( y = e^{-3x} \) over the interval \([0, 1]\), we use the formula for the surface area of a solid of revolution. The formula is:

\[
A = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]

Given \( y = e^{-3x} \), we first need to find \( \frac{dy}{dx} \).

\[
\frac{dy}{dx} = \frac{d}{dx}(e^{-3x}) = -3e^{-3x}
\]

Next, we substitute \( y \) and \( \frac{dy}{dx} \) into the formula for the surface area:

\[
A = 2\pi \int_{0}^{1} e^{-3x} \sqrt{1 + (-3e^{-3x})^2} \, dx
\]

Simplify the integrand:

\[
A = 2\pi \int_{0}^{1} e^{-3x} \sqrt{1 + 9e^{-6x}} \, dx
\]

To solve this integral, we'll use substitution and numerical methods. Let’s first set up the integral for numerical evaluation.

\[
\int_{0}^{1} e^{-3x} \sqrt{1 + 9e^{-6x}} \, dx
\]

This integral doesn't have a straightforward antiderivative, so we'll approximate its value numerically.

Using a numerical integration method (like Simpson's Rule or a numerical integration calculator), we can evaluate the integral. Let’s compute this using Simpson's Rule for better approximation:

1. Divide the interval \([0, 1]\) into \( n \) subintervals (we'll use \( n = 4 \) for simplicity).

2. Calculate the function values at these points and apply Simpson's Rule.

\[ 
I \approx \frac{h}{3} \left[ f(0) + 4f\left(\frac{1}{4}\right) + 2f\left(\frac{1}{2}\right) + 4f\left(\frac{3}{4}\right) + f(1) \right]
\]

Where \( f(x) = e^{-3x} \sqrt{1 + 9e^{-6x}} \) and \( h = \frac{1 - 0}{4} = 0.25 \).

Calculating the function values:

\[
f(0) = e^{0} \sqrt{1 + 9e^{0}} = \sqrt{10}
\]
\[
f\left(\frac{1}{4}\right) = e^{-0.75} \sqrt{1 + 9e^{-1.5}} \approx 0.472 \sqrt{1 + 9(0.223)} \approx 0.472 \sqrt{2.007} \approx 0.665
\]
\[
f\left(\frac{1}{2}\right) = e^{-1.5} \sqrt{1 + 9e^{-3}} \approx 0.223 \sqrt{1 + 9(0.050)} \approx 0.223 \sqrt{1.45} \approx 0.269
\]
\[
f\left(\frac{3}{4}\right) = e^{-2.25} \sqrt{1 + 9e^{-4.5}} \approx 0.105 \sqrt{1 + 9(0.011)} \approx 0.105 \sqrt{1.099} \approx 0.110
\]
\[
f(1) = e^{-3} \sqrt{1 + 9e^{-6}} \approx 0.050 \sqrt{1 + 9(0.002)} \approx 0.050 \sqrt{1.018} \approx 0.050
\]

Putting it all together:

\[
I \approx \frac{0.25}{3} \left[ \sqrt{10} + 4(0.665) + 2(0.269) + 4(0.110) + 0.050 \right]
\]

\[
I \approx \frac{0.25}{3} \left[ 3.162 + 2.660 + 0.538 + 0.440 + 0.050 \right]
\]

\[
I \approx \frac{0.25}{3} \left[ 6.850 \right]
\]

\[
I \approx 0.571
\]

Finally, multiply by \( 2\pi \):

\[
A \approx 2\pi \times 0.571 \approx 3.587 \text{ square units}
\]

So, the surface area of the revolution is approximately \( 3.587 \) square units.

Would you like to know more details or have any questions?

Here are 8 relative questions that expand on the previous information:
1. What is the definition of the surface area of revolution?
2. How do you determine the surface area for other curves?
3. What are some common methods for numerical integration?
4. How does Simpson's Rule work for numerical integration?
5. What is the significance of the derivative in the surface area formula?
6. Can you provide an example of surface area of revolution about the y-axis?
7. How do you handle surface area calculations for more complex functions?
8. What are the limitations of numerical methods for integration?

**Tip:** For complex integrals that do not have straightforward antiderivatives, numerical methods like Simpson's Rule or numerical integration tools are very useful.

Learn how to compute the surface area of revolution about the x-axis for the function y = e^-3x over the interval [0, 1]. Detailed step-by-step solution using calculus and numerical methods. Suitable for college-level mathematics.

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