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as level p1 👍👍👍 integration?
plz show the full method
4 Answers
- Anonymous1 year agoFavorite Answer
V = pi [integrate x^2 dy , with y=A to y=1 ]
integrate (8/y^2-2)^2 dy
= 64/y^4-32/y^2+4) dy
= -320/y^5 + (32/3)/y^3 +4y
V = pi [-320/y^5 + (32/3)/y^3 +4y, y=A to y=1]
= pi [-320/A^5 + (32/3)/A^3 +4A - (-320+(32/3)+4)]
= pi [-320/A^5 + (32/3)/A^3 +4A + 305.33]
- az_lenderLv 71 year ago
V = integral from y = 1 to 2 of
pi x^2 dy
= integral from y = 1 to 2 of
pi(8y^(-2) - 2)^2 dy
= integral from y = 1 to 2 of
pi(64y^(-4) - 32y^(-2) + 4) dy.
After integration you have
pi[(-64/3)y^(-3) + 32y^(-1) + 4y]
Plug in the limits and you get
pi[(-64/3)(1/8) + 32*(1/2) + 8 - (-64/3)(1) - 32 - 4]
= pi[(64/3)(7/8) - 16 + 4] = 20pi/3.
- VamanLv 71 year ago
integrate the term
2 pi x dy= 2 pi (8/y^2-1) dy
2 pi (-1/3 y^(-3) -y) Put the limits 1 and A
1pi(-1/3 (1/A^3)-A)- (-1/3-1)= 2 pi (4/3- 1/(3A^3)-A).
This seem to be volume. Please check it.
