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

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To find the area of the surface generated by rotating the curve about the y-axis, we can use the formula for the surface area of revolution, which is given by:

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

Given the curve ( x = t + 1 ) and ( y = \frac{1}{2}t^2 + t ), with ( 0 \leq t \leq 2 ), we need to find ( \frac{dx}{dt} ) and then substitute it into the formula.

First, calculate ( \frac{dx}{dt} ):

[ \frac{dx}{dt} = \frac{d}{dt}(t + 1) = 1 ]

Now, we substitute ( \frac{dx}{dt} = 1 ) and ( y = \frac{1}{2}t^2 + t ) into the formula:

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

Simplify the expression:

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

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

Now integrate term by term:

[ A = \sqrt{2}\pi \left[\frac{1}{3}t^3 + t^2\right]_{0}^{2} ]

[ A = \sqrt{2}\pi \left[\frac{1}{3}(2)^3 + (2)^2 - \left(\frac{1}{3}(0)^3 + (0)^2\right)\right] ]

[ A = \sqrt{2}\pi \left[\frac{8}{3} + 4\right] ]

[ A = \sqrt{2}\pi \left[\frac{8}{3} + \frac{12}{3}\right] ]

[ A = \sqrt{2}\pi \cdot \frac{20}{3} ]

[ A = \frac{20\sqrt{2}\pi}{3} ]

Therefore, the area of the surface generated by rotating the curve about the y-axis is ( \frac{20\sqrt{2}\pi}{3} ).

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