Автор - Murрhу

Вычислить интеграл. Подробно, с графиком...

Ответ

Автор - hello93

iiintlimits_{D}{xyz}dxdydz=iiintlimits_{D}pcos(phi)sin(theta)psin(phi)sin(theta)pcos(theta)p^2sin(theta)dpdphi dtheta=iiintlimits_{D}p^5sin^3(theta)cos(theta)sin(phi)cos(phi)dpdphi dtheta=intlimits_0^{pi/2}{sin(phi)cos(phi)}dphiintlimits_0^{pi/2}{sin^3(theta)cos(theta)dtheta}intlimits_0^4p^5dp=intlimits_0^{pi/2}{sin(phi)cos(phi)}dphiintlimits_0^{pi/2}{sin^3(theta)cos(theta)dtheta}timesBig(dfrac{p^6}{6}Big)Big|^4_0=

=dfrac{2048}{3}intlimits_0^{pi/2}{sin(phi)cos(phi)}dphiintlimits_0^{pi/2}{sin^3(theta)cos(theta)dtheta}=begin{vmatrix}u=sin(theta)\du=cos(theta)dtheta\a=0 , b=1end{vmatrix}=

=dfrac{2048}{3}intlimits_0^{pi/2}sin(phi)cos(phi)dphiintlimits_0^{1}u^3du=dfrac{2048}{12}intlimits_0^{pi/2}sin(phi)cos(phi)dphi=begin{vmatrix}t=cos(phi)\dt=-sin(phi)dphi\a=11, b=0end{vmatrix}=-dfrac{2048}{12}intlimits_1^0{t}dt=dfrac{2048}{12}intlimits_0^1{t}dt=dfrac{2048}{12}dfrac{t^2}{2}Big|^1_0=dfrac{2048}{24}=dfrac{256}{3}approx85.(3)

График - сфера в первой четверти

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