VC5.50

From Exampleproblems

Evaluate with the help of divergence theorm the integral $\iint_S[xz^2 dy dz+(x^2y-z^3)dz dx+(2xy+y^2z)dx dy]\,$ , where S is the entire surface of the hemispherical region bounded by $z=\sqrt{(a^2-x^2-y^2)},z=0\,$ .

Let V be the volume enclosed by S.Hence S is the point of the sphere given.

By divergence theorem, $\iint_S[xz^2dy dz+(x^2y-z^3)dz dx+(2xy+y^2z)dx dy]\,$

= $\iiint_V[\frac{\partial}{\partial x}(xz^2)+\frac{\partial}{\partial y}(x^2y-z^3)+\frac{\partial}{\partial z}(2xy+y^2z)]dV\,$

= $\iiint_V(z^2+x^2+y^2)dV\,$ --(1)

To simplify the triple integral,we use the spherical polar coordinates $(r,\theta,\phi)\,$ so that $x=r\sin\theta\cos\phi,y=r\sin\theta\sin\phi,z=r\cos\theta,dV=(dr)(r d\theta)(r\sin\theta d\phi)=r^2\sin\theta dr d\theta d\phi\,$ --(2)

In order to cover the volume,the limits are r=0 to a $\theta=0,\frac{\pi}{2},\phi=0,2\pi\,$

Here $\iiint_V(x^2+y^2+z^2)dV\,$

= $\int_{r=0}^{a}\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{2\pi} r^2\cdot r^2\sin\theta dr d\theta d\phi\,$

= $\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{2\pi}[\frac{r^5}{5}]_0^a \sin\theta d\theta d\phi\,$

= $\frac{a^5}{5}\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{2\pi}\sin\theta d\theta d\phi\,$

= $\frac{a^5}{5}\int_{\theta=0}^{\frac{\pi}{2}}\sin\theta[\phi]_{0}^{2\pi} d\theta\,$

= $\frac{2\pi a^5}{5}\int_{\theta=0}^{\frac{\pi}{2}}\sin\theta d\theta\,$

= $\frac{2\pi a^5}{5}[-\cos\theta]_{0}^{\frac{\pi}{2}}=\frac{2\pi a^5}{5}\,$ --(3)

Using (3) in (1),we see that the value of the given surface integral is $\frac{2\pi a^5}{5}\,$

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