91Ó°ÊÓ

Find \(\operatorname{div}(\operatorname{curl} \mathbf{F})=\nabla \cdot(\nabla \times \mathbf{F})\) $$ \mathbf{F}(x, y, z)=x^{2} z \mathbf{i}-2 x z \mathbf{j}+y z \mathbf{k} $$

Prove the property for vector fields \(\mathbf{F}\) and \(\mathbf{G}\) and scalar function \(f .\) (Assume that the required partial derivatives are continuous.) $$ \operatorname{curl}(\mathbf{F}+\mathbf{G})=\operatorname{curl} \mathbf{F}+\operatorname{curl} \mathbf{G} $$

Find the area of the lateral surface (see figure) over the curve \(C\) in the \(x y\) -plane and under the surface \(z=f(x, y),\) where Lateral surface area \(=\int_{C} f(x, y) d s\) \(f(x, y)=y, \quad C:\) line from (0,0) to (4,4)

Prove the property for vector fields \(\mathbf{F}\) and \(\mathbf{G}\) and scalar function \(f .\) (Assume that the required partial derivatives are continuous.) $$ \operatorname{curl}(\nabla f)=\nabla \times(\nabla f)=\mathbf{0} $$

Find the area of the lateral surface (see figure) over the curve \(C\) in the \(x y\) -plane and under the surface \(z=f(x, y),\) where Lateral surface area \(=\int_{C} f(x, y) d s\) \(f(x, y)=x y, \quad C: x^{2}+y^{2}=1\) from (1,0) to (0,1)

Prove the property for vector fields \(\mathbf{F}\) and \(\mathbf{G}\) and scalar function \(f .\) (Assume that the required partial derivatives are continuous.) $$ \operatorname{div}(\mathbf{F}+\mathbf{G})=\operatorname{div} \mathbf{F}+\operatorname{div} \mathbf{G} $$

Find the area of the lateral surface (see figure) over the curve \(C\) in the \(x y\) -plane and under the surface \(z=f(x, y),\) where Lateral surface area \(=\int_{C} f(x, y) d s\) \(f(x, y)=y+1, \quad C: y=1-x^{2}\) from (1,0) to (0,1)

Prove the property for vector fields \(\mathbf{F}\) and \(\mathbf{G}\) and scalar function \(f .\) (Assume that the required partial derivatives are continuous.) $$ \operatorname{div}(\mathbf{F} \times \mathbf{G})=(\operatorname{curl} \mathbf{F}) \cdot \mathbf{G}-\mathbf{F} \cdot(\operatorname{curl} \mathbf{G}) $$

Find the area of the lateral surface (see figure) over the curve \(C\) in the \(x y\) -plane and under the surface \(z=f(x, y),\) where Lateral surface area \(=\int_{C} f(x, y) d s\) \(f(x, y)=x y, \quad C: y=1-x^{2}\) from (1,0) to (0,1)

Find the area of the lateral surface (see figure) over the curve \(C\) in the \(x y\) -plane and under the surface \(z=f(x, y),\) where Lateral surface area \(=\int_{C} f(x, y) d s\) \(f(x, y)=x^{2}-y^{2}+4, \quad C: x^{2}+y^{2}=4\)