MHB Solids of Revolution - Negative Volume

[Dethrone]

Hmm... can you redo the following integral with a couple more steps? (Wondering)
$$\int_{-1}^{1} (x^2+y^2)dx$$

I had no idea what to do, but I thought of y^2 as a constant since we're integrating with respect to x.

$$= [\frac{x^3}{3}]_{-1}^1=\frac{2}{3}$$

EDIT: Somehow, I thought the integral of a constant was 0. Let me re-work this.

 

[I like Serena]

I had no idea what to do, but I thought of y^2 as a constant since we're integrating with respect to x.

$$= [\frac{x^3}{3}]_{-1}^1$$

Yep! Let's treat $y^2$ as a constant.
Let's suppose $y^2=2$ for a minute, so it really looks like a constant.
What is:
$$\int 2\,dx$$

 

[Dethrone]

Yep! Let's treat $y^2$ as a constant.
Let's suppose $y^2=2$ for a minute, so it really looks like a constant.
What is:
$$\int 2\,dx$$

It's $2x$. I think I can do this now. I knew it was a constant, so I got excited when I was doing it, but I really have no clue why I thought the integral of a constant was 0...

 

[I like Serena]

It's $2x$. I think I can do this now. I knew it was a constant, so I got excited when I was doing it, but I really have no clue why I thought the integral of a constant was 0...

Oh, I know.
The derivative of a constant is 0.
It means that the anti-derivative of 0 is a constant, which is the other way around.
I suspect you were just mixing them up, which is a pretty normal thing to do.

 

[Dethrone]

$$V=\int_{-1}^{1}\int_{-1}^{1} x^2+y^2 \,dx \,dy$$
$$=\int_{-1}^{1} [\frac{x^3}{3}+xy^2]_{-1}^1\,dy$$
$$=\int_{-1}^{1}\frac{2}{3}+2y^2\,dy$$
$$=\frac{2}{3}[y+y^3]_{-1}^1$$
$$=\frac{8}{3} units^3$$

:D :D :D

 

[I like Serena]

$$V=\int_{-1}^{1}\int_{-1}^{1} x^2+y^2 \,dx \,dy$$
$$=\int_{-1}^{1} [\frac{x^3}{3}+xy^2]_{-1}^1\,dy$$
$$=\int_{-1}^{1}\frac{2}{3}+2y^2\,dy$$
$$=\frac{2}{3}[y+y^3]_{-1}^1$$
$$=\frac{8}{3} units^3$$

:D :D :D

Yep! (Emo)

 

[Dethrone]

Yup, using advice from this thread, I can now solve most types of these questions. In particular, I'm happy to have gotten this one right: Find the volume of the frustum of a cone whose lower base is of radius R, upper base is of radius r, and altitude is h.

I also want to point out, although not another "method", but some questions require the use of this formula (which is kind of obvious): (I guess you can say the disc/washer method uses this too, technically)

$$V= \int A(x) dx$$

"A solid has a circular base of radius 4 units. Find the volume of the solid if every plane perpendicular to the fixed diameter is an isosceles right triangle with the hypotenuse in the plane of the base".