Solids of Revolution - Negative Volume

In summary, the problem was that the height of the shells I created was all negative, so the volume was negative. I fixed it by including an absolute value in my integration.
  • #36
Rido12 said:
$$V=\int_{-1}^{1}\int_{-1}^{1} x^2+y^2 \,dx \,dy$$
$$=\int_{-1}^{1} [\frac{x^3}{3}+xy^2]_{-1}^1\,dy$$
$$=\int_{-1}^{1}\frac{2}{3}+2y^2\,dy$$
$$=\frac{2}{3}[y+y^3]_{-1}^1$$
$$=\frac{8}{3} units^3$$

:D :D :D

Yep! (Emo)
 
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  • #37
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".
 
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