I'm having some troubles setting this calculus problem. I need to find the volume and the area of the solid resulting from the rotation about the $x$ axis of the figure given by
- the $x$-axis
- the curve $y^2 = x + 1$ for $-1 \leq x \leq 0$ and $y \geq 0$.
- the semicircle given by the equation $x^2 + (y - \frac 12)^2 = \frac 14$ for $x \geq 0.$
For the volume given by the rotation of $y^2 = x + 1$, I solved for $y$ and got:
$V_1 = \pi\int_{-1}^0 \sqrt{x + 1}^2 dx = \frac \pi2$.
For the volume given by the rotation of the semicircle I also solved for y but I get this rather ugly integral to evaluate:
$V_2 = \pi\int_0^\frac 12 (\frac 12 + \sqrt{\frac 14 + x^2})^2 dx$ wich I don't know how to solve. My guess is that the resulting volument is $V = V_1 + V_2$. Is my work correct so far? How do I continue?
Your work for $V_1$ looks fine, but for $V_2$ I think you want
$\displaystyle V_2=\int_0^{1/2}\pi\left((R(x))^2-(r(x))^2\right)dx=\int_0^{1/2}\pi\left(\bigg(\frac{1}{2}+\sqrt{\frac{1}{4}-x^2}\bigg)^2-\bigg(\frac{1}{2}-\sqrt{\frac{1}{4}-x^2}\bigg)^2\right)dx$
$\displaystyle\hspace{.2 in}=\int_0^{1/2}\pi\left(2\sqrt{\frac{1}{4}-x^2}\right)dx=2\pi\int_0^{1/2}\sqrt{\frac{1}{4}-x^2}dx$.
Now you can use the fact that this integral represents $\frac{1}{4}$ of the area of a circle, or use trig substitution.