How do you find the area of the surface generated by rotating the curve about the y-axis #x=2t+1, y=4-t, 0<=t<=4#?

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

The formula is given by:

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

In this case, the curve is given parametrically by ( x = 2t + 1 ) and ( y = 4 - t ), where ( 0 \leq t \leq 4 ).

First, express ( y ) in terms of ( t ), then differentiate ( x ) with respect to ( t ) to find ( \frac{{dx}}{{dt}} ), and differentiate ( y ) with respect to ( t ) to find ( \frac{{dy}}{{dt}} ).

Next, substitute these expressions into the formula for ( A ) and integrate over the given range of ( t ), which is ( 0 ) to ( 4 ).

[ A = \int_{0}^{4} 2\pi (4 - t) \sqrt{1 + \left(\frac{{dx}}{{dt}}\right)^2} , dt ]

[ A = \int_{0}^{4} 2\pi (4 - t) \sqrt{1 + \left(\frac{{dx}}{{dt}}\right)^2} , dt ]

[ A = \int_{0}^{4} 2\pi (4 - t) \sqrt{1 + \left(\frac{{d(2t + 1)}}{{dt}}\right)^2} , dt ]

[ A = \int_{0}^{4} 2\pi (4 - t) \sqrt{1 + 4^2} , dt ]

[ A = \int_{0}^{4} 2\pi (4 - t) \sqrt{17} , dt ]

Now, integrate this expression with respect to ( t ) from ( 0 ) to ( 4 ) to find the surface area.