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

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

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Answer 1 · 2017-07-17T19:32:54

First we will combine this 2 equation to find x in term of y and then we will calculate the area.

Explanation:

$y = 4 - t \Leftrightarrow t = 4 - y$ $y=4-t iff t=4-y$
we plug this value into $x = 2 t + 1 \Leftrightarrow x = 2 (4 - y) + 1 \Leftrightarrow x = 9 - 2 y$ $x=2t+1 iff x=2(4-y)+1 iff x=9-2y$
this gives us the curve
graph{x=9-2y [-10, 10, -5, 5]}
and $0 \leq y \leq 4$ $0<=y<=4$
the area generated is given by the integral:
$E = \int_{0}^{4} (9 - 2 y) d y$ $E=int_0^4(9-2y)dy$
because we rotate the curve about the y-axis
so $E = {[9 y - y^{2}]}_{0}^{2} = 36 - 16 = 20$ $E=[9y-y^2]_0^4=36-16=20$

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

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Explanation:

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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<=4x=2t+1,y=4−t,0≤t≤4x=2t+1, y=4-t, 0<=t<=4?

1 Answer

Explanation:

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