Let $A_{R,r}$ be a set with $0\lt r\lt R$ $$A_{R,r}= \{(x,y,z)\in\mathbb R^3\mid (R-\sqrt{x^2+y^2})^2+z^2\le r^2\}$$ Evaluate $\lambda_3(A_{R,r})$
Answer: I used Cavalieri's principle and set $$A_z=\{(x,y)\in\mathbb R^2\mid R-\sqrt{r^2-z^2}\le\sqrt{x^2+y^2}\le R+\sqrt{r^2-z^2}\}$$ and then follows $$\lambda_3(A_{R,r})=\int_{-r}^{r} \lambda_2(A_z)dz$$ with $$\lambda_2(A_z)=\int_{R-\sqrt{r^2-z^2}}^{R+\sqrt{r^2-z^2}} \lambda_1(B_x^z)dx$$ where $$B_x^z=\{y\in\mathbb R\mid (R-\sqrt{r^2-z^2})^2-x^2\le y^2\le (R+\sqrt{r^2-z^2})^2-x^2\}$$ hence $$\lambda_1(B_x^z)=\int_{\sqrt{(R-\sqrt{r^2-z^2})^2-x^2}}^{\sqrt{(R+\sqrt{r^2-z^2})^2-x^2}} dy=\sqrt{(R+\sqrt{r^2-z^2})^2-x^2}-\sqrt{(R-\sqrt{r^2-z^2})^2-x^2}$$