Given a vector \[\overline {\rm{u}} = \frac{1}{3}\left( { - {{\rm{y}}^3}{\rm{\hat i}} + {{\rm{x}}^3}{\rm{\hat j}} + {{\rm{z}}^3}{\rm{\hat k}}} \right)\] and \[{{\rm{\hat n}}}\] as the unit normal vector to the surface of the hemisphere (x2 + y2 + z2 = 1; z ≥ 0), the value of integral \[\int {\left( {\nabla \times \overline {\rm{u}} } \right) \cdot {\rm{\hat n}}} {\rm{dS}}\] evaluated on the curved surface of the hemisphere S is
A. \[\pi \]
B. \[\frac{\pi }{2}\]
C. \[\frac{{ - \pi }}{2}\]
D. \[\frac{\pi }{3}\]
Answer: Option B
Related Questions on Calculus
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. \[\infty \]
C. \[\frac{1}{2}\]
D. \[ - \infty \]
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
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