# What is the surface area produced by rotating #f(x)=1/e^(x^2), x in [-1,1]# around the x-axis?

Just like the volume of a shape is found by summing up each infinitesimal cross section of area, the surface area is found by summing up each infinitesimal cross section of perimeter.

Therefore, our integral becomes:

Therefore, our integral changes to:

This is a good stopping point, unless the problem wants the decimal form of the answer, in which case plugging this into a calculator gives:

Final Answer

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To find the surface area produced by rotating (f(x) = \frac{1}{e^{x^2}}) on the interval ([-1, 1]) around the x-axis, we can use the formula for the surface area of a curve rotated around the x-axis:

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

First, we find ( \frac{dy}{dx} ):

[ f(x) = \frac{1}{e^{x^2}} ]

[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{e^{x^2}} \right) = -2x \cdot \frac{1}{e^{x^2}} ]

Next, we find ( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} ):

[ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \left( \frac{-2x}{e^{x^2}} \right)^2} = \sqrt{1 + \frac{4x^2}{e^{2x^2}}} ]

Now, we can calculate the surface area:

[ S = 2\pi \int_{-1}^{1} \frac{1}{e^{x^2}} \sqrt{1 + \frac{4x^2}{e^{2x^2}}} , dx ]

This integral does not have a simple closed-form solution and would typically be evaluated numerically.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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