Surface area of solid of revolution (no calculus)

[V0ODO0CH1LD]

Homework Statement

Consider the region of the x y plane given by the inequality:

x^2 + 4x + y^2 - 4x - 8 ≤ 0;

If this region rotates an angle of π/6 radians around the line given by the equation x + y = 0, it will create a solid of revolution with surface area equal to

(i) (128/3)π; (ii) (128/4)π; (iii) (128/5)π; (iv) (128/6)π; (v) (128/7)π

Homework Equations

area of a sphere = 4πr^2

The Attempt at a Solution

okay, so first I factored the inequality into:

(x + 2)^2 + (y - 2)^2 ≤ 16;

Which means it's a circle centered at (2, -2) and with r = 4. The line x + y = 0 goes through the center of the circle. Now I am stuck.. cause I though that if it rotated an angle of π radians around the line it would give the full external area of the sphere so if it rotated only π/6 it would give 1/6 of it. Which would equal to (4π4^2)/6 = (64/6)π. But that is not one of the answers..

 

[LCKurtz]

I agree with everything you have said. My best guess is that either the problem meant for you to calculate its volume instead of surface area, or whoever wrote the answer choices accidentally did that instead of calculating the surface area. The volume does give one of those answers.

 

[haruspex]

a solid of revolution with surface area equal to

It's a solid, not a shell. Have you considered all surfaces of the solid?

 

[V0ODO0CH1LD]

It's a solid, not a shell. Have you considered all surfaces of the solid?

I haven't though about that! Of course it's not just a shell! It's an inequality that generates the solid of revolution! In that case I would have the (64/6)π + 2(π4^2) one area of the circle for where the solid is being generated from and one from where the solid is rotating to! And that equals (128/3)π!

Thanks!

EDIT: Wait, is it a solid not a "shell" because they specified by saying so, or is it a solid not a shell because they gave the thing to generate the solid to be an inequality?

 

[LCKurtz]

I haven't though about that! Of course it's not just a shell! It's an inequality that generates the solid of revolution! In that case I would have the (64/6)π + 2(π4^2) one area of the circle for where the solid is being generated from and one from where the solid is rotating to! And that equals (128/3)π!

Thanks!

EDIT: Wait, is it a solid not a "shell" because they specified by saying so, or is it a solid not a shell because they gave the thing to generate the solid to be an inequality?

Interestingly enough, that is also the answer to its volume.

 

[aralbrec]

EDIT: Wait, is it a solid not a "shell" because they specified by saying so, or is it a solid not a shell because they gave the thing to generate the solid to be an inequality?

The inequality gives a filled in circle and rotated results in a solid no matter what they say :)