Register
Find the area of the surface generated by revolving x = t y = t^2 1 ≤ t ≤ 2 around the y-axis
184k views
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
6.3k points

2 Answers

5 votes
answer is 4^22x (sorry it’s late)
User Cybercop
by
5.0k points
5 votes

Answer:

4^22x

Explanation:

User RapGodRory
by
5.2k points
Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

5.9m questions

7.7m answers

Other Questions

Solve for y in terms of x. 2/3y - 4 = x y = x + 6 y = -x + 4 y = -x + 6 y = x + 4
Aliyah has $24 to spend on seven pencils. After buying them she had $10. How much did each pencil cost?
Use the normal model ​n(11371137​,9696​) for the weights of steers. ​a) what weight represents the 3737thth ​percentile? ​b) what weight represents the 9999thth ​percentile? ​c) what's the iqr of the weights
Evaluate a + b when a = –16.2 and b = –11.4 A. –27.6 B. –4.8 C. 4.8 D. 27.6
What is the value of x? (-4) - x = 5
Link Copied!