Answer to Question #223030 in Calculus for Hanna

Question

Calculate the volume created by revolution around the axes OX, of D of the plane (OXY) where D={(x,y) ∈R: 0 ≤ y ≤x, (x-2)² +y² ≤ 4}

Answer 1

Solution.

$$V_x=\pi \int_a^b f^2(x)dx$$

So, we will have

$$V_x=V_1+V_2=\pi \int_0^2 x^2dx+\pi\int_2^4 (\sqrt{4-(x-2)^2})^2dx$$

$\int_0^2 x^2dx=\frac{x^3}{3}|_0^2=\frac{8}{3}$

$\int_2^4 (4-x^2+4x-4)dx=\int_2^4 (-x^2+4x)dx=\newline (-\frac{x^3}{3}+4\frac{x^2}{2})|_2^4=-\frac{64}{3}+2\cdot 16+\frac{8}{3}-2\cdot 4=\newline -\frac{56}{3}+24$

From here

$V_x=(\frac{8}{3}-\frac{56}{3}+24)\pi=(24-\frac{48}{3})\pi=(24-16)\pi=8\pi$ sq.units.

Answer. $8\pi$ sq.units.