Volume of Solid of Revolution for y=x^2-2, y=0 about y=-1

[californicate]

Homework Statement

Find the volume of the solid of revolution obtained by rotating the area bounded by the curves about the line indicated.
y=x²-2, y=0 about y=-1. Need only consider part above y=-1

Homework Equations

V=∏_a∫^b[f(x)]²dx

The Attempt at a Solution

I'm mainly unsure of my solution, as it gives me an answer but I feel that my bounds aren't selected properly.
Roots: -√2 to √2
V=∏_-√2 ∫^√2[(x-1)²-(-1)²]dx
I get an answer of 21.91 u³, but as I'm working in the negative y I feel like the y upper and y lower I've selected aren't correct as the parabola is not above the line. Should I integrate in terms of x instead?

 

[haruspex]

Did you sketch the region? Looks to me like a horizontal cylinder, axis y=-1, with a dimple in the right hand end (positive x end). That means the bounds for y depend on x. Split the range of x into two integrals.

 

[LCKurtz]

Homework Statement

Find the volume of the solid of revolution obtained by rotating the area bounded by the curves about the line indicated.
y=x²-2, y=0 about y=-1. Need only consider part above y=-1

Homework Equations

V=∏_a∫^b[f(x)]²dx

The Attempt at a Solution

I'm mainly unsure of my solution, as it gives me an answer but I feel that my bounds aren't selected properly.
Roots: -√2 to √2
V=∏_-√2 ∫^√2[(x-1)²-(-1)²]dx
I get an answer of 21.91 u³, but as I'm working in the negative y I feel like the y upper and y lower I've selected aren't correct as the parabola is not above the line. Should I integrate in terms of x instead?

Do you mean should you integrate in terms of y instead? I would say a $dy$ integral would be easier. Use shells. As Haruspex suggests, draw a picture if you haven't already.

 

[californicate]

So, if integrating in terms of y, I would integrate x=sqrt(y+2) from -1 to 0?, So my final integral is pi (integral from -1 to 0) (sqrt(y+2))^2 dy?

 

[haruspex]

So, if integrating in terms of y, I would integrate x=sqrt(y+2) from -1 to 0?, So my final integral is pi (integral from -1 to 0) (sqrt(y+2))^2 dy?

No. This is the cylinder method. Each cylinder has length x, but what is its surface area?

 

[californicate]

We didn't use terms like by shells and by cylinders in lecture so I'm trying to follow along, but since if I'm doing by cylinders would I then require the constant to be 2pi?

 

[haruspex]

We didn't use terms like by shells and by cylinders in lecture so I'm trying to follow along, but since if I'm doing by cylinders would I then require the constant to be 2pi?

Yes, there's a 2π in there, but that's not the only error in your integral.
Your cylindrical element has y=-1 as axis. It is dy thick. What are its length and circumference?

 

[californicate]

It's radius is 1, so it's circumference is 2pi. It's length is 2x, so 2sqrt(y+2)?

 

[LCKurtz]

It's radius is 1, so it's circumference is 2pi. It's length is 2x, so 2sqrt(y+2)?

That's right for the length of the $dy$ element. But it is at a variable y-coordinate $y$ and it is being rotated about $y=-1$ so its radius of rotation isn't $1$ and its circumference isn't $2\pi$. It depends on $y$.