Find the volume of the solid region $T$ enclosed by $S$
Let $S$ be the surface obtained by rotating the curve $$ x=cos (t), y=0, z=sin (2t), -frac{pi}{2} leq t leq frac{pi}{2} $$
about the $z$- axis. Find the volume of the solid region $T$ enclosed by $S$.
Answer:
By divergence theorem,
the volume of the solid region $T$ is $ iiint_T 1 dV=iint_S F cdot n dS$, where $F$ is any vector field with $div(F)=1$.
i.e., $V=iint_S F cdot n dS$.
Now we have to define a parametrization $r( theta, t)$ for $S$, where $ theta$ is taken from cylindrical coordinates.
But I can not do this.
Here vector field $F$ can be taken arbitrarily so that $div(F)=1$.
we can choose $F=frac{1}{2} leftlangle x,0,z rightrangle$, then $div(F)=1$.
If we can find the parametrization $r( theta,t)$ of $S$, then we can find the normal $n $ on S.
Please help me to evaluate the integral
multivariable-calculus divergence
asked Nov 25 '18 at 22:58
arifamath
1176
add a comment |
Let $S$ be the surface obtained by rotating the curve $$ x=cos (t), y=0, z=sin (2t), -frac{pi}{2} leq t leq frac{pi}{2} $$
about the $z$- axis. Find the volume of the solid region $T$ enclosed by $S$.
Answer:
By divergence theorem,
the volume of the solid region $T$ is $ iiint_T 1 dV=iint_S F cdot n dS$, where $F$ is any vector field with $div(F)=1$.
i.e., $V=iint_S F cdot n dS$.
Now we have to define a parametrization $r( theta, t)$ for $S$, where $ theta$ is taken from cylindrical coordinates.
But I can not do this.
Here vector field $F$ can be taken arbitrarily so that $div(F)=1$.
we can choose $F=frac{1}{2} leftlangle x,0,z rightrangle$, then $div(F)=1$.
If we can find the parametrization $r( theta,t)$ of $S$, then we can find the normal $n $ on S.
Please help me to evaluate the integral
multivariable-calculus divergence
asked Nov 25 '18 at 22:58
arifamath
1176
add a comment |
Let $S$ be the surface obtained by rotating the curve $$ x=cos (t), y=0, z=sin (2t), -frac{pi}{2} leq t leq frac{pi}{2} $$
about the $z$- axis. Find the volume of the solid region $T$ enclosed by $S$.
Answer:
By divergence theorem,
the volume of the solid region $T$ is $ iiint_T 1 dV=iint_S F cdot n dS$, where $F$ is any vector field with $div(F)=1$.
i.e., $V=iint_S F cdot n dS$.
Now we have to define a parametrization $r( theta, t)$ for $S$, where $ theta$ is taken from cylindrical coordinates.
But I can not do this.
Here vector field $F$ can be taken arbitrarily so that $div(F)=1$.
we can choose $F=frac{1}{2} leftlangle x,0,z rightrangle$, then $div(F)=1$.
If we can find the parametrization $r( theta,t)$ of $S$, then we can find the normal $n $ on S.
Please help me to evaluate the integral
multivariable-calculus divergence
asked Nov 25 '18 at 22:58
arifamath
1176
Let $S$ be the surface obtained by rotating the curve $$ x=cos (t), y=0, z=sin (2t), -frac{pi}{2} leq t leq frac{pi}{2} $$
about the $z$- axis. Find the volume of the solid region $T$ enclosed by $S$.
Answer:
By divergence theorem,
the volume of the solid region $T$ is $ iiint_T 1 dV=iint_S F cdot n dS$, where $F$ is any vector field with $div(F)=1$.
i.e., $V=iint_S F cdot n dS$.
Now we have to define a parametrization $r( theta, t)$ for $S$, where $ theta$ is taken from cylindrical coordinates.
But I can not do this.
Here vector field $F$ can be taken arbitrarily so that $div(F)=1$.
we can choose $F=frac{1}{2} leftlangle x,0,z rightrangle$, then $div(F)=1$.
If we can find the parametrization $r( theta,t)$ of $S$, then we can find the normal $n $ on S.
Please help me to evaluate the integral
multivariable-calculus divergence
multivariable-calculus divergence
asked Nov 25 '18 at 22:58
arifamath
1176
asked Nov 25 '18 at 22:58
arifamath
1176
edited Nov 25 '18 at 23:05
asked Nov 25 '18 at 22:58
arifamath
1176
asked Nov 25 '18 at 22:58
arifamath
1176
asked Nov 25 '18 at 22:58
arifamath
1176
1176
add a comment |
add a comment |
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