<s>[INST] Computing volume using shell resulted in negative value? The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...
Use cylindrical shells to compute the volume of $y=x^2$ and $y=2-x^2$, revolved about $x=2$
$$x^2=2-x^2→2x^2=2→\frac{2x^2}2=\frac{2}2→x^2=1$$
$$x=1, x=-1$$
Radius is $$r=2-x$$
Height is $$x^2-(2-x^2)$$
Finding the Integral,
$$\int_{\neg1}^12π(2-x)(x^2-(2-x^2 ))dx=2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)+C$$
Finding the limits,
$$\lim_{x→-1+}⁡2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)=2π(\frac{4(-1)^3}3-4(-1)-\frac{2(-1)^4}3+2(-1)^2-(-1)^2+\frac{(-1)^4}6)=19.897$$
$$\lim_{x→1-}⁡2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)=2π(\frac{4(1)^3}3-4(1)-\frac{2(1)^4}3+2(1)^2-(1)^2+\frac{(1)^4}6)=-13.614$$
And finally, computing the volume from the obtained limits
$$V=-13.614-19.897=-33.510$$
So, if someone could let me know if (or where) I went wrong, that would be great.
 [/INST] In the region $-1 \leq x \leq1$, $2-x^2 > x^2$ (the $2-x^2$ curve is above the $x^2$ curve). So the height is $(2-x^2)-x^2$, not $x^2-(2-x^2)$.
This fix should simply change the sign of your answer -- which is exactly the issue you wanted to address!
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