Question

Вычислить интеграл. Подробно, с графиком...

Answer 1

$\iiint\limits_{D}{xyz}dxdydz=\iiint\limits_{D}p\cos(\phi)\sin(\theta)p\sin(\phi)\sin(\theta)p\cos(theta)p^2\sin(\theta)dpd\phi d\theta=\iiint\limits_{D}p^5\sin^3(\theta)cos(\theta)\sin(\phi)\cos(\phi)dpd\phi d\theta=\int\limits_0^{\pi/2}{\sin(\phi)\cos(\phi)}d\phi\int\limits_0^{\pi/2}{sin^3(\theta)cos(\theta)d\theta}\int\limits_0^4p^5dp=\int\limits_0^{\pi/2}{\sin(\phi)\cos(\phi)}d\phi\int\limits_0^{\pi/2}{sin^3(\theta)cos(\theta)d\theta}\times\Big(\dfrac{p^6}{6}\Big)\Big|^4_0=$

$=\dfrac{2048}{3}\int\limits_0^{\pi/2}{\sin(\phi)\cos(\phi)}d\phi\int\limits_0^{\pi/2}{sin^3(\theta)cos(\theta)d\theta}=\begin{vmatrix}u=\sin(\theta)\\du=\cos(\theta)d\theta\\a=0 , b=1\end{vmatrix}=$

$=\dfrac{2048}{3}\int\limits_0^{\pi/2}\sin(\phi)\cos(\phi)d\phi\int\limits_0^{1}u^3du=\dfrac{2048}{12}\int\limits_0^{\pi/2}\sin(\phi)\cos(\phi)d\phi=\begin{vmatrix}t=\cos(\phi)\\dt=-sin(\phi)d\phi\\a=11, b=0\end{vmatrix}=-\dfrac{2048}{12}\int\limits_1^0{t}dt=\dfrac{2048}{12}\int\limits_0^1{t}dt=\dfrac{2048}{12}\dfrac{t^2}{2}\Big|^1_0=\dfrac{2048}{24}=\dfrac{256}{3}\approx85.(3)$

График - сфера в первой четверти