1). Find the area of the upper part of

${x}^{2} + {y}^{2} = 4$

12). Find the volume if the area of no 1 is rotated
about X axis

1). Find the area of the upper part of

Jawab:

Penjelasan dengan langkah-langkah:

$1.\\x^2+y^2=4\\y^2=4-x^2\\y=sqrt{4-x}\\\L=intlimits^2_{-2} {sqrt{4-x^2}} , dx~~~~~~~~~~let~sinalpha =frac{x}{2}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2sinalpha =x\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2cosalpha dalpha =dx\\L=intlimits^2_{-2} {sqrt{4-(2sinalpha )^2}} , 2cosalpha dalpha \\L=2intlimits^2_{-2} {cosalpha sqrt{4cos^2alpha }} , dalpha \\L=2intlimits^2_{-2} {2cos^2alpha } , dalpha \\L=2intlimits^2_{-2} {1+cos2alpha } , dalpha$

$\L=2(alpha +frac{1}{2}sin2alpha)|^2_{-2}\\L=2arcsinfrac{x}{2}+2.frac{x}{2}.frac{sqrt{4-x^2}}{2}|^2_{-2}\\L=2arcsinfrac{2}{2}+2.frac{2}{2}.frac{sqrt{4-2^2}}{2}-2arcsinfrac{-2}{2}+2frac{-2}{2}.frac{sqrt{4-(-2)^2}}{2}\\L=2.frac{pi}{2}+0-2(-frac{pi}{2})+0\\L=2pi$

$12.\V=pi intlimits^2_{-2} {(sqrt{4-x^2})^2} , dx\\V=pi intlimits^2_{-2} {4-x^2} , dx\\V=pi (4x-frac{1}{3}x^3)|^2_{-2}\\V=pi (4.2-frac{1}{3}(2)^3-4(-2)+frac{1}{3}(-2)^3))\\V=pi (8-frac{8}{3}+8-frac{8}{3})\\V=frac{32}{3}pi$