Answer to Question #104270 in Calculus for Ajay

Question #104270
∫∫∫cos(x^2+y^2+z^2)^(3/2) dxdydz
1
Expert's answer
2020-03-03T11:44:06-0500

cos(x2+y2+z2)3/2dxdydz\iiint{\cos(x^2+y^2+z^2)^{3/2} dxdydz}


Convert x,y,zx,y,z to spherical coordinates r,φ,θ.r, \varphi, \theta.

x=rcos(φ)sin(θ)y=rsin(φ)sin(θ)z=rcos(θ)x=r\cos(\varphi)\sin(\theta) \\ y=r\sin(\varphi)\sin(\theta) \\ z=r\cos(\theta)


The absolute value of Jacobian is r2sin(θ).r^2\sin(\theta). Then the original integral is



cos((r2)3/2)rsin(θ)drdφdθ=r2cos(r3)sin(θ)drdφdθ=r2cos(r3)drdφsin(θ)dθ=13cos(r3)d(r3)dφsin(θ)dθ=13φsin(r3)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)}

We have omitted the constants for convenience.

Return to x,y,z.x,y,z.

r=x2+y2+z2φ=arctan(yx)θ=arccos(zx2+y2+z2)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)}


Therefore, the original integral is


cos(x2+y2+z2)3/2dxdydz=13arctan(yx)sin((x2+y2+z2)3/2)cos[arccos(zx2+y2+z2)]=13arctan(yx)sin((x2+y2+z2)3/2)zx2+y2+z2\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}}



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