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Calculus IIIChapter 6: Surface IntegralsSection 6-1

Curl and Divergence β€” Question 1

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Question 1

For 𝑭(x,y,z)=⟨x2y,yz2,xz⟩,\mathbf F(x,y,z)=\langle x^2y,\,yz^2,\,xz\rangle, compute βˆ‡β‹…π‘­\nabla\cdot\mathbf F and βˆ‡Γ—π‘­\nabla\times\mathbf F.

Tasks

  1. Compute the divergence term by term.

  2. Compute all three components of the curl.

  3. Evaluate both quantities at (1,2,βˆ’1)(1,2,-1).

Original worksheet page 1: question and worked solution for 6-1-001
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Question 1 – Solution

Strategy. Label the components P,Q,RP,Q,R and apply the divergence and curl formulas in a fixed order.

Step 1: Divergence With P=x2yP=x^2y, Q=yz2Q=yz^2, and R=xzR=xz, βˆ‡β‹…π‘­=Px+Qy+Rz=2xy+z2+x.\nabla\cdot\mathbf F=P_x+Q_y+R_z =2xy+z^2+x. Thus βˆ‡β‹…π‘­=2xy+z2+x.\boxed{\nabla\cdot\mathbf F=2xy+z^2+x}.

Step 2: Curl βˆ‡Γ—π‘­=⟨Ryβˆ’Qz,Pzβˆ’Rx,Qxβˆ’Py⟩=⟨0βˆ’2yz,0βˆ’z,0βˆ’x2⟩.\begin{align*} \nabla\times\mathbf F &=\langle R_y-Q_z,\,P_z-R_x,\,Q_x-P_y\rangle\\ &=\langle 0-2yz,\,0-z,\,0-x^2\rangle. \end{align*} Therefore βˆ‡Γ—π‘­=βŸ¨βˆ’2yz,βˆ’z,βˆ’x2⟩.\boxed{\nabla\times\mathbf F=\langle-2yz,-z,-x^2\rangle}.

Step 3: Evaluate At (1,2,βˆ’1)(1,2,-1), βˆ‡β‹…π‘­=6,βˆ‡Γ—π‘­=⟨4,1,βˆ’1⟩.\boxed{\nabla\cdot\mathbf F=6}, \qquad \boxed{\nabla\times\mathbf F=\langle 4,1,-1\rangle}.

Verification The curl components follow the cyclic order (Ryβˆ’Qz,Pzβˆ’Rx,Qxβˆ’Py)(R_y-Q_z,P_z-R_x,Q_x-P_y); recomputing from the determinant form gives the same signs.

Original worksheet page 2: question and worked solution for 6-1-001

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