# How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #y=e^x# on the interval #0<=x<=1# ?

Since you are only given a single function and we are rotating about the axis of the parameter, this requires the disk method. The disk method is:

We have the known values:

And now we can substitute:

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To find the surface area of the solid obtained by rotating the region bounded by ( y = e^x ) on the interval ( 0 \leq x \leq 1 ) about the x-axis, you would use the formula for surface area of revolution, which is ( S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} , dx ), where ( f(x) ) represents the function describing the curve and ( [a, b] ) is the interval of integration.

In this case, ( f(x) = e^x ), so ( f'(x) = e^x ). Thus, the integral becomes ( S = 2\pi \int_{0}^{1} e^x \sqrt{1 + e^{2x}} , dx ).

This integral represents the surface area of the solid of revolution. By evaluating this integral using appropriate techniques, such as substitution or integration by parts, you can find the surface area of the solid obtained by rotating the given region about the x-axis.

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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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