# How do you find the area of the surface generated by rotating the curve about the x-axis #x=t^2+t, y=2t+1, 0<=t<=1#?

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

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

Where (y) is the function representing the curve, and (a) and (b) are the limits of integration.

First, find (\frac{dy}{dx}) by differentiating (y) with respect to (x). Then, substitute (y), (\frac{dy}{dx}), and the limits of integration into the formula and integrate with respect to (x) from the lower limit to the upper limit.

Given the curve (x = t^2 + t) and (y = 2t + 1) with (0 \leq t \leq 1), differentiate (y) with respect to (x) to find (\frac{dy}{dx}).

[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} ]

[ \frac{dy}{dt} = 2 ] [ \frac{dx}{dt} = 2t + 1 ]

Substitute these into the formula:

[ A = \int_{0}^{1} 2\pi (2t + 1) \sqrt{1+(2)^2} , dt ]

[ A = \int_{0}^{1} 2\pi (2t + 1) \sqrt{5} , dt ]

[ A = 2\pi\sqrt{5} \int_{0}^{1} (2t + 1) , dt ]

[ A = 2\pi\sqrt{5} \left[\frac{t^2}{2} + t\right]_0^1 ]

[ A = 2\pi\sqrt{5} \left[\left(\frac{1}{2} + 1\right) - \left(0\right)\right] ]

[ A = 2\pi\sqrt{5} \left(\frac{3}{2}\right) ]

[ A = 3\pi\sqrt{5} ]

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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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