What is the surface area of the solid created by revolving #f(x)=-sqrtx/e^x# over #x in [0,1]# around the x-axis?

Answer 1

I got as far as the definite integral but then I got stuck at the solution of the integral....probably there is a mistake or there is an easier way...

I considered the function and the rotation:

then I used the standard technique:

giving the expression of the surface area #S# as:

#S=int_0^1(2pi)(-sqrt(x)/e^x)*sqrt(1+(4x^2-4x+1)/(4e^(2x)*x))dx#

which I am not able to solve!!!

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Answer 2

To find the surface area of the solid created by revolving the function f(x) = -√x/e^x over the interval [0,1] around the x-axis, you can use the formula for the surface area of revolution:

SA = ∫[a,b] 2πf(x)√(1 + (f'(x))^2) dx

In this case, the interval is [0,1], and the function is f(x) = -√x/e^x. You'll need to find the derivative f'(x) and then plug into the formula to calculate the surface area.

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Answer from HIX Tutor

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.

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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