find the volume generated by rotating the region bounded by $y^2 - x^2 = 1$ and y = 2 about the x - axis

i got $2\sqrt 3 \pi$ but my book says $4 \sqrt 3 \pi$ o.O

Question

find the volume generated by rotating the region bounded by $y^2 - x^2 = 1$ and y = 2 about the x - axis

i got $2\sqrt 3 \pi$ but my book says $4 \sqrt 3 \pi$ o.O

lgbasallote · Answer

the equation i used is $$2\pi \int_0^2 y(\sqrt{y^2 - 1}) dy$$

dumbcow · Answer

using disk or washer method would be easier here

cross-sections are rings with outer radius of 2, inner radius of y
$$V = \pi \int\limits_{-\sqrt{3}}^{\sqrt{3}}2^{2} -(\sqrt{1+x^{2}})^{2} dx = \pi \int\limits_{-\sqrt{3}}^{\sqrt{3}}(3-x^{2}) dx$$

dumbcow · Answer

oh i just noticed something, in the limits of your integral....it should be from 1 to 2 not 0 to 2

lgbasallote · Answer

why 1??

lgbasallote · Answer

oh wait...it's the graph isnt it

dumbcow · Answer

because 1 is vertex of hyperbola

dumbcow · Answer

(0,1) i should say

lgbasallote · Answer

did that come from the drawing of the graph? i drew it wrong?

lgbasallote · Answer

oh yeah lol i drew it wrong :p

lgbasallote · Answer

well the answer to my equation is still $2\pi$

lgbasallote · Answer

i mean $2\sqrt 3 \pi$

dumbcow · Answer

ahh i got it...the height should be 2x or 2sqrt(y^2-1)

lgbasallote · Answer

but whyyyy T_T

dumbcow · Answer

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