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Find the area of the surface generated by revolving x = t y = t^2 1 ≤ t ≤ 2 around the y-axis

Find the area of the surface generated by revolving x = t y = t^2 1 ≤ t ≤ 2 around the y-axis
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3 votes
Find the area of the surface generated by revolving x = t y = t^2 1 ≤ t ≤ 2 around the y-axis

User Naeemgik
by
8.8k points

2 Answers

5 votes
answer is 4^22x (sorry it’s late)
User Cybercop
by
8.1k points
5 votes

Answer:

4^22x

Explanation:

User RapGodRory
by
8.0k points
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