∭ cos ( x 2 + y 2 + z 2 ) 3 / 2 d x d y d z \iiint{\cos(x^2+y^2+z^2)^{3/2} dxdydz} ∭ cos ( x 2 + y 2 + z 2 ) 3/2 d x d y d z
Convert x , y , z x,y,z x , y , z r , φ , θ . r, \varphi, \theta. r , φ , θ .
x = r cos ( φ ) sin ( θ ) y = r sin ( φ ) sin ( θ ) z = r cos ( θ ) x=r\cos(\varphi)\sin(\theta) \\
y=r\sin(\varphi)\sin(\theta) \\
z=r\cos(\theta) x = r cos ( φ ) sin ( θ ) y = r sin ( φ ) sin ( θ ) z = r cos ( θ )
The absolute value of Jacobian is r 2 sin ( θ ) . r^2\sin(\theta). r 2 sin ( θ ) .
∭ cos ( ( r 2 ) 3 / 2 ) ⋅ r sin ( θ ) d r d φ d θ = ∭ r 2 cos ( r 3 ) sin ( θ ) d r d φ d θ = ∫ r 2 cos ( r 3 ) d r ∫ d φ ∫ sin ( θ ) d θ = 1 3 ∫ cos ( r 3 ) d ( r 3 ) ∫ d φ ∫ sin ( θ ) d θ = − 1 3 φ sin ( r 3 ) cos ( θ ) \iiint {\cos((r^2)^{3/2})\cdot r\sin(\theta)dr d\varphi d\theta} = \iiint{r^2\cos(r^3)\sin(\theta)dr d\varphi d\theta } \\
= \int{r^2\cos(r^3)dr}\int{d\varphi}\int{\sin(\theta)d\theta} = \frac{1}{3}\int{\cos(r^3)d(r^3)}\int{d\varphi}\int{\sin(\theta)d\theta} \\
= -\frac{1}{3}\varphi\sin(r^3)\cos{(\theta)} ∭ cos (( r 2 ) 3/2 ) ⋅ r sin ( θ ) d r d φ d θ = ∭ r 2 cos ( r 3 ) sin ( θ ) d r d φ d θ = ∫ r 2 cos ( r 3 ) d r ∫ d φ ∫ sin ( θ ) d θ = 3 1 ∫ cos ( r 3 ) d ( r 3 ) ∫ d φ ∫ sin ( θ ) d θ = − 3 1 φ sin ( r 3 ) cos ( θ ) We have omitted the constants for convenience.
Return to x , y , z . x,y,z. x , y , z .
r = x 2 + y 2 + z 2 φ = arctan ( y x ) θ = arccos ( z x 2 + y 2 + z 2 ) r=\sqrt{x^2+y^2+z^2} \\
\varphi = \arctan{\left( \frac{y}{x}\right)} \\
\theta = \arccos{\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)} r = x 2 + y 2 + z 2 φ = arctan ( x y ) θ = arccos ( x 2 + y 2 + z 2 z )
Therefore, the original integral is
∭ cos ( x 2 + y 2 + z 2 ) 3 / 2 d x d y d z = − 1 3 arctan ( y x ) sin ( ( x 2 + y 2 + z 2 ) 3 / 2 ) cos [ arccos ( z x 2 + y 2 + z 2 ) ] = − 1 3 arctan ( y x ) sin ( ( x 2 + y 2 + z 2 ) 3 / 2 ) z x 2 + y 2 + z 2 \iiint{\cos(x^2+y^2+z^2)^{3/2} dxdydz} \\
= -\frac{1}{3} \arctan{\left( \frac{y}{x}\right)}\sin((x^2+y^2+z^2)^{3/2})\cos{\left[ \arccos{\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)} \right]} \\
= -\frac{1}{3} \arctan{\left( \frac{y}{x}\right)}\sin((x^2+y^2+z^2)^{3/2})\frac{z}{\sqrt{x^2+y^2+z^2}} ∭ cos ( x 2 + y 2 + z 2 ) 3/2 d x d y d z = − 3 1 arctan ( x y ) sin (( x 2 + y 2 + z 2 ) 3/2 ) cos [ arccos ( x 2 + y 2 + z 2 z ) ] = − 3 1 arctan ( x y ) sin (( x 2 + y 2 + z 2 ) 3/2 ) x 2 + y 2 + z 2 z
#340153 on Dec 2023
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