Green's theorem in the plane relates
Answer: A
The theorem states
\[ \oint_{C}\left(M\,dx + N\,dy\right) \;=\; \iint_{R}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)dA. \]
The other pairings describe Gauss's divergence theorem and Stokes's theorem, both of which are three-dimensional generalisations of this plane result
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By Green's theorem, the area of a region \( R \) equals
Answer: D
Take \( M = -y \) and \( N = x \), so that \( \partial N/\partial x - \partial M/\partial y = 1 + 1 = 2 \):
\[ \oint_{C}\left(x\,dy - y\,dx\right) \;=\; \iint_{R} 2\,dA \;=\; 2A. \]
Hence the half. This is how a planimeter measures an area from its boundary alone, without ever sampling the interior
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Stokes's theorem states that the circulation of a field round a closed curve equals
Answer: C
The theorem reads
\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} \;=\; \iint_{S} (\nabla \times \mathbf{F}) \cdot \mathbf{n}\,dS. \]
That the surface may be any one spanning the curve follows because the curl is solenoidal, so its flux depends only on the boundary. It also explains at once why an irrotational field has zero circulation
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Gauss's divergence theorem states that the flux out of a closed surface equals
Answer: B
The theorem reads
\[ \oiint_{S} \mathbf{F} \cdot \mathbf{n}\,dS \;=\; \iiint_{V} \nabla \cdot \mathbf{F}\,dV. \]
It is the statement that what leaves a region is what the sources inside it produce, and it is what turns the integral form of a conservation law into its differential form. A closed surface has no boundary curve, so no line integral arises
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For \( \mathbf{F} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \), the flux out of any closed surface enclosing volume \( V \) is
Answer: A
The divergence is constant:
\[ \nabla \cdot \mathbf{F} \;=\; 1 + 1 + 1 \;=\; 3, \]
so the divergence theorem gives \( \iiint 3\,dV = 3V \) whatever the shape of the surface. For a sphere of radius \( a \) this is \( 3 \times \tfrac{4}{3}\pi a^{3} = 4\pi a^{3} \), matching the direct calculation
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The circulation of an irrotational field round any closed curve in a simply connected region is
Answer: D
By Stokes's theorem the circulation equals the flux of the curl through a spanning surface, and an irrotational field has zero curl:
\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} \;=\; \iint_{S} \mathbf{0} \cdot \mathbf{n}\,dS \;=\; 0. \]
Simple connectedness matters: on a region with a hole the spanning surface may not exist, and the circulation can then be non-zero
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Match the theorem with the pair of integrals it connects. 1. Green 2. Stokes 3. Gauss divergence P. Closed line integral and flux of a curl Q. Closed plane line integral and area integral R. Closed surface flux and volume integral
Answer: C
Green's theorem is the plane case, trading a boundary line integral for an integral over the enclosed area. Stokes lifts it into space, the area integral becoming a flux of the curl through a spanning surface. Gauss relates a closed surface to the volume it encloses, and has no boundary curve at all
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Match the property with the field it describes. 1. Solenoidal 2. Irrotational 3. Both at once P. Expressible as a gradient Q. Expressible as a curl R. Its potential satisfies Laplace's equation
Answer: B
A divergence-free field is the curl of a vector potential, since every curl is solenoidal. A curl-free field is the gradient of a scalar, since every gradient is irrotational. A field with both properties has \( \nabla \cdot \nabla\phi = 0 \), so its scalar potential is harmonic
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Put these steps of testing a plane field for conservativeness and finding its potential in order. 1. Integrate one component to obtain a candidate potential 2. Compare \( \partial N/\partial x \) with \( \partial M/\partial y \) 3. Differentiate the candidate in the other variable and match the remaining component 4. Write the field as \( M\mathbf{i} + N\mathbf{j} \)
Answer: A
The components are identified first, then the cross-derivative test decides whether a potential exists at all - there is no point integrating if it fails. One component is then integrated, leaving a function of the other variable, which the second component fixes. Integrating before testing is what wastes the effort
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Consider the following statements about vector fields. 1. The curl of a gradient vanishes identically 2. A solenoidal field has a vector potential 3. Every field with zero curl has zero divergence
Answer: D
The first two are standard. The third confuses the two conditions: the radial field \( x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \) is irrotational yet has divergence three. The two properties are independent, and a field satisfying both is the special harmonic case
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