Question

Consider the following assertions:
(I) \(\displaystyle\iint_S(x\hat i+y\hat j+z\hat k)\cdot\hat n\,dS=2V\), where \(\hat n\) is the unit outward surface normal and S is a closed surface enclosing a volume V.
(II) \(\displaystyle\int_C(yz\hat i+xz\hat j+xy\hat k)\cdot d\vec r=3\), where C is the boundary of a surface S.
Choose the correct option:

• A. (I) is correct, but (II) is false
• B. (I) is false, but (II) is correct
• C. Both (I) and (II) are false
• D. Both (I) and (II) are correct

Hint:

Apply the divergence theorem to (I) and recognize (II) as a gradient field.

Answer 1

The Correct Option is C

Step 1: Check (I) using the divergence theorem, \(\displaystyle\iint_S\vec F\cdot\hat n\,dS=\iiint_V(\nabla\cdot\vec F)\,dV\). For \(\vec F=(x,y,z)\), \(\nabla\cdot\vec F=1+1+1=3\), so \[\iint_S(x,y,z)\cdot\hat n\,dS=\iiint_V 3\,dV=3V.\] This equals \(3V\), not \(2V\), so statement (I) is false.

Step 2: Check (II). Notice \(yz,\ xz,\ xy\) are the partial derivatives of \(\varphi=xyz\), so \(\vec G=(yz,xz,xy)=\nabla(xyz)\) is a gradient field.

Step 3: Since C is the boundary of a surface S, C is a closed curve. The line integral of a gradient field around any closed curve is zero: \[\oint_C\nabla\varphi\cdot d\vec r=\varphi(\text{end})-\varphi(\text{start})=0\] because the start and end points coincide on a closed loop. So the integral equals 0, not 3, and statement (II) is false.

Step 4: Both (I) and (II) are false. \[\boxed{\text{Both (I) and (II) are false}}\]