# How do you find the exact area of the surface obtained rotating the curve about the #x#-axis of #y=sqrt(8-x)#, #2<=x<=8#?

graph{y = sqrt(8-x) [-1.07, 10.97, -1.25, 5.76]} Rotating about x-axis:

Arc length is:

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To find the exact area of the surface obtained by rotating the curve ( y = \sqrt{8 - x} ) about the x-axis over the interval ( 2 \leq x \leq 8 ), you can use the formula for the surface area of revolution, which is ( S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ). Substituting ( y = \sqrt{8 - x} ), you can find ( \frac{dy}{dx} ). Then, plug these values into the formula and evaluate the integral from ( x = 2 ) to ( x = 8 ) to find the surface 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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