TJC H2 Math question

E7E72 SG

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.

wee_ws SG

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

E7E72 SG

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?

wee_ws SG

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