We have a hyperbola : (x+6)^2/5 -(y-1)^2=1
It intersects x=1 at (1,3) and (1,-1)
How do i find the volume when the region enclosed by x=1 and the hyperbola is rotated completely about the y-axis?
I have attempted to apply pi (integral with boundary 3,1) hyperbola eqn in terms of x^2 dy, to no avail as i am unable to isolate the x^2 term. Also, if i use the quadratic equation on x, then square it, i appear to be getting an answer way off at 1000+ units.
PS: correct answer is 84.7 if i remember correctly.
Many thanks.
Hi,
I believe there is an error in the equation of the hyperbola, it should have been
(x-6)^2 / 5 - (y-1)^2 = 1.
Volume = pi integral {-1, 3} of {6-sqrt(5+5(y-1)^2} dx - 4 pi.
The answer is 87.7 via GC.
Thanks!
Cheers,
Wen Shih
Nice! I assume you found the x^2 term via the -b+-sqrtb2-4ac/2 formula, but how do you get to rejecting the 6+sqrt(5+(5y-1)^2 answer?
Hi,
I just made x the subject in the equation.
If you draw the hyperbola, one branch lies on the left of x = 6 and the other branch lies on the right of x = 6. For this question, we are interested only in the left branch.
Thanks.
Cheers,
Wen Shih