Finding volume when rotating y = 0.5x² around x = -3 using shell method

[Kqwert]

Homework Statement

A container with height $$4.5$$ is created by rotating the curve $$y = 0.5x^2$$ $$0 \leq x \leq 3$$ around $$x = -3$$ and putting a plane bottom in the box. Find the volume $$V$$ of the box.

Homework Equations

The Attempt at a Solution

I want to solve this by using the shell method. I have put up the following integral, which will be the volume of the revolution. It is however not correct, and I haven't really used the information about the height given in the text.. could anyone help me?

$$2\pi\int_0^3 (x+3)(0.5x^2) \, dx$$

 

[mjc123]

You are omitting the central cylinder (between x = 0 and -6).
Personally I think it would be easier to take circular slices in the y direction.

 

[Kqwert]

How do you go about calculating the volume of the central cylinder?

 

[Charles Link]

I think you just edited it with the $2 \pi$, and it looks correct to me.

 

[Kqwert]

I think you just edited it with the $2 \pi$, and it looks correct to me.

No, the post haven't been edited. I think it´s true what mjc123 said about omitting the central cylinder. I am however not exactly sure how to visualize it.

 

[Charles Link]

No, the post haven't been edited. I think it´s true what mjc123 said about omitting the central cylinder. I am however not exactly sure how to visualize it.

I am a little puzzled by this one also, because it doesn't look like a box. $\\$ Edit: I think I see what they are wanting, but I'll let you work on it for a few minutes.

 

[Kqwert]

I am a little puzzled by this one also, because it doesn't look like a box.

I´ve translated it from my own language, so box is probably the wrong description. Let's say container instead. (edited first post)

 

[mjc123]

Oh, and you're also integrating under the curve, i.e. outside the box. Replace 0.5x² by (4.5 - 0.5x²).
Try drawing a diagram of this to see what the "box" looks like (it's like a bowl).

 

[mjc123]

This is what I think you're being asked to find the volume of. What is the volume of the bowl?

 

[Kqwert]

Oh, and you're also integrating under the curve, i.e. outside the box. Replace 0.5x² by (4.5 - 0.5x²).
Try drawing a diagram of this to see what the "box" looks like (it's like a bowl).

Is it like this? We want to find the volume which results when rotating the gray colored area around x = -3, but we also need to include the "inner" volume which I´ve colored orange. The volume of this orange part should be pi*3^2*4.5..?

 

[mjc123]

Looks like you've got it.

 

[Kqwert]

Looks like you've got it.

I still get wrong. My answer is (135/2)*pi. Is it the same as you?

 

[mjc123]

No, I get 114.75*pi, by both methods.

 

[Kqwert]

No, I get 114.75*pi, by both methods.

Excellent, had forgotten that the graph was 0.5x^2 and not x^2... got it now!