Find the volume of the solid formed when the area bounded by x^2-y^2=9, y=x+5, and the x axis is rotated about x=1.

Answer 1

Looks like a small trianglulish area. Curve intersects x-axis at x = -3.
Line intersects at x = -5
Line intersects curve at (-17/5,8/5)

Using discs,

v = ∫pi(R^2-r^2) dy
where R^2 = (1 - (5-y))^2 = (y-4)^2
r^2 = 9+y^2

v = pi*∫[0,8/5](y-4)^2-(9+y^2) dy
= pi*∫[0,8/5](-8y+7) dy
= pi*(-4y^2+7y)[0,8/5]
= 0.96 pi = 3.016

Answer 2

What a messy question!!!

first find the intersection of the hyperbola and the straight line,
I got (-3.4 , 1.6)

from x^2 - y^2=9 ---> x = ±(9-y^2)^(1/2)
then the inner radius = 1 + √(9-y^2)
and (inner radius)^2 = (1+2√(9-y^2) + 9 - y^2
= 10 - y^2 + 2√(9-y^2)

outer radius = 1 - (y-5) = 6-y
(outer radius)^2 = 36 - 12y + y^2

volume = π∫(36-12y+y^2 - (10 - y^2 + 2√(9-y^2)) dy from 0 to 1.6

= π∫( 26 - 12y + 2y^2 + 2√(9-y^2) ) dy from 0 to 1.6

The messiest part is to integrate √9-y^2)

Let's do that separately.....
(after all that below, I noticed that I used x instead of y, so all that below should be in terms of y )</b?
let x = 3sinθ

dx = 3cosθ dθ

∫ √(9 - x²) dx
= ∫ √(9 - (3sinθ)²) 3cosθ dθ
= 3∫ √(9 - 9sin²θ) cosθ dθ
= 9∫√(1 - sin²θ) cosθ dθ
= 9∫√(cos²θ) cosθ dθ
= 9∫cos²θ dθ
= 9/2*∫(1 + cos(2θ)) dθ
= 9/2*(θ + sin(2θ) / 2 ) + C
= 9/2*(θ + sinθcosθ ) + C

Now:
x = 3sinθ
so
x/3 = sinθ
θ = sin ֿ ¹ (x/3)
θ = cosֿ ¹ (√(9 - x²)/3)

so

= 9/2*(θ + sinθcosθ ) + C
= 9/2*(sin ֿ ¹ (x/3) + sin(sin ֿ ¹ (x/3))cos(cosֿ ¹ (√(9 - x²)/3)) ) + C
= 9/2*(sin ֿ ¹ (x/3) + x/3* √(9 - x²)/3 ) + C
= 9/2*sin ֿ ¹ (x/3) + x√(9 - x²)/2 + C

OK, your turn
Sure hope I did not mess up

Answer 3

looking at your

r^2 = 9+y^2

you are rotating it about the y-axis, not x=1

Answer 4

Ouch! Got me!