What is the surface area of the solid created by revolving #f(x) = xe^(2x) , x in [2,3]# around the x axis?

Answer 1

Volume : #(61e^12--23e^8)/32pi# = 30974.5 cubic units, nearly. The graph is not to scale. I would try to get surface area, soon.

graph{x^2e^(2x) [-5, 5, -10000, 10000]}

The volume = #pi int f^2 dx#, with x from 2 to 3
#=pi int x^2e^(4x )dx#, with x from 2 to 3

Now,

# int x^2e^(4x) dx#
#= 1/4int x^2d(e^(4x))#

#=1/4(x^2e^(4x)--int e^(4x)d(x^2))

#=1/4x^2e^4x-1/8 intxd(e^(4x))#
#=1/4x^2e^4x-1/8(xe^(4x)-inte^(4x) dx))#
#=1/4x^2e^4x-1/8xe^(4x)+1/32e^(4x) + C#

Substituting the limits for the difference, the value here is

#(61e^12--23e^8)/32#.

So, the volume is

#(61e^12--23e^8)/32pi# = 30974.5 cubic units, nearly.
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Answer 2

To find the surface area of the solid created by revolving the function ( f(x) = xe^{2x} ) around the x-axis over the interval ( x \in [2,3] ), we can use the formula for surface area of a solid of revolution:

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

where ( y = f(x) ) and ( \frac{dy}{dx} ) is the derivative of ( f(x) ) with respect to ( x ).

First, let's find ( \frac{dy}{dx} ):

[ f(x) = xe^{2x} ] [ \frac{dy}{dx} = e^{2x} + 2xe^{2x} ]

Now, we can plug these values into the formula and integrate over the interval ( x \in [2,3] ):

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

After integrating this expression, we obtain the surface area of the solid of revolution.

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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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