# How do you find the area of the surface generated by rotating the curve about the y-axis #y=1/4x^4+1/8x^-2, 1<=x<=2#?

By Power Rule,

So, the arc length element is:

#sqrt{1+(dy/dx)^2}=sqrt(1+(x^3)^2-1/2+(x^(-3)/4)^2) =sqrt((x^3)^2+1/2+(x^(-3)/4)^2)=sqrt((x^3+x^(-3)/4)^2) =x^3+x^(-3)/4#

Hence, the surface area can be expressed as:

#S=2pi int_1^2 x sqrt(1+(dy/dx)^2) dx =2pi int_1^2(x^4+x^(-2)/4) dx#

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To find the area of the surface generated by rotating the curve about the y-axis, you can use the formula:

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

Where (a) and (b) are the limits of integration. In this case, (a = 1) and (b = 2).

First, find ( \frac{dy}{dx} ) by differentiating the equation ( y = \frac{1}{4}x^4 + \frac{1}{8}x^{-2} ) with respect to (x).

[ \frac{dy}{dx} = x^3 - \frac{1}{4}x^{-3} ]

Now, plug (y) and ( \frac{dy}{dx} ) into the formula and integrate from (a = 1) to (b = 2).

[ A = 2\pi \int_{1}^{2} x \sqrt{1 + \left(x^3 - \frac{1}{4}x^{-3}\right)^2} dx ]

Then solve the integral to find the area.

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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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