How would I set up an integral for the volume of the solid bounded between these two curves:
$$y=x$$ $$y=\frac{2x}{1+x^3}$$
Rotated about x=-1.
And these two curves:
$$y^2 - x^2 = 1$$ $$y=2$$ About y-axis.
Which method would be most useful in this situation? Shell, washer or disc? I'm having a hard time visualizing this right now.
First Pair: The graph of $y=2x/(1+x^{3})$ looks like $y=2x$ for $x \approx 0$, and resembles $2/x^{2}$ for large positive and negative $x$. $y=2x/(1+x^{3})$ has a vertical asymptote at $x=-1$, and is well-behaved everywhere else. So $y=2x/(1+x^{3})$ starts out above $y=x$ for $x > 0$, but that switches as the $x^{3}$ dominates and pulls $y=2x/(1+x^{3})$ toward $0$ for large $x$. The two graphs intersect at $x=1$ as can be seen by solving $$ \frac{2x}{1+x^{3}}=x, x > 0,\;\;\; \implies x=1. $$ The region between these graphs lies between $x=0$ and $x=1$. You want to rotate the region around $x=-1$. It's easiest to compute the height of the region as a function of $x$: $$ \mbox{height}(x) = \frac{2x}{1+x^{3}}-x,\;\;\; 0 \le x \le 1,\\ \mbox{radius}(x) = x+1. $$ The volume of this region is $$ V = \int_{0}^{1} [2\pi\cdot\mbox{radius}(x)]\mbox{height}(x)\,dx \\ = 2\pi\int_{0}^{1}2(x+1)\left[\frac{2x}{1+x^{3}}-x\right]\,dx $$
Second Pair: You should recognize $y^{2}-x^{2}=1$ as the graph of a hyperbola, the two branches of which can be expressed as $y=\pm \sqrt{1+x^{2}}$. The vertices of these branches are at $x=0,\;y=\pm 1$. Only the upper branch intersects $y=2$. You are asked to rotate this region of intersection about the $y$ axis. This is one probably easiest using disks because the radius is $$\mbox{radius}=x=\sqrt{y^{2}-1},\;\; 1 \le y \le 2,$$ which has a square that is easily parameterized by $y$. The volume integral is $$ V = \pi \int_{1}^{2}(y^{2}-1)\,dy. $$