Answer to Question #223286 in Calculus for irvon tyuiol

Question #223286

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}



1
Expert's answer
2021-10-26T02:34:31-0400
Vx=πabf2(x)dxV_x=\pi \int_a^b f^2(x)dx

So, we will have



Vx=V1+V2=π02x2dx+π24(4(x2)2)2dxV_x=V_1+V_2=\pi \int_0^2 x^2dx+\pi\int_2^4 (\sqrt{4-(x-2)^2})^2dx

02x2dx=x3302=83\int_0^2 x^2dx=\frac{x^3}{3}|_0^2=\frac{8}{3}

24(4x2+4x4)dx=24(x2+4x)dx=(x33+4x22)24=643+216+8324=563+24\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

Vx=(83563+24)π=(24483)π=(2416)π=8πV_x=(\frac{8}{3}-\frac{56}{3}+24)\pi=(24-\frac{48}{3})\pi=(24-16)\pi=8\pi sq.units.


Answer. 8π8\pi sq.units.



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