Using disk method in integration

Question

I need to find the volume of the region defined by the inequalities $$\begin{align*}y & <\frac 12\\0< y & <1-x^2\end{align*}$$revolved around the $x$ axis. I tried separating them and then integrating from zero to $0.707$ for $y<\frac 12$ and then from $0.707$ to one for $0<y<1-x^2$ but without a correct result.

Answer 1

The region is simply the area enclosed by the graphs $y=1-x^2$, $y=\tfrac 12$, and the $x$ axis. Since the area is symmetric, consider only the region lying in the first quadrant. We'll double our answer to arrive at the area.

First, let's find the intersection of the two functions$$1-x^2=\frac 12\qquad\implies\qquad x=\frac 1{\sqrt2}$$

The region to the left of $x$ is simply a rectangle whose area is simply$$A_{\text{rectangle}}=\frac 1{\sqrt2}\times\frac 12=\frac 1{2\sqrt2}$$The area of the region to the right of $x$ can be done by drawing a "slice" from the function to the $x$ axis. The height of that is simply$$r=1-x^2$$Therefore, rotating it about the $x$ axis and integrating, we get that the area of the right - hand side is simply$$A_{\text{right}}=\pi\int\limits_{\tfrac 1{\sqrt2}}^1\mathrm dx\,(1-x^2)^2+\pi\int\limits_0^{\frac 1{\sqrt2}}\frac {\mathrm dx}4$$ Hence, the total area is$$A\color{blue}{=2\pi\int\limits_{\frac 1{\sqrt2}}^1\mathrm dx\,(1-x^2)^2+2\pi\int\limits_0^{\frac 1{\sqrt2}}\frac {\mathrm dx}4}$$

Answer 2

For $0\to1/\sqrt2$, you get$$\pi\int_0^{1/\sqrt2}(1/2)^2dx$$For $1/\sqrt2\to1$, you get$$\pi\int_{1/\sqrt2}^1(1-x^2)^2dx$$The final answer is double the sum of the above integrals due to symmetry.