If $\vec{F}=-4 \vec{G}$ and S is an oriented surface, then $\int _ { S } \vec { F } \cdot \vec { d A } = - 4 \int _ { S } \vec { G } \cdot \vec { d A }$ .

Question

Quizplus · Accepted Answer

The property of linearity of the surface integral allows us to factor out constants from the integral, so multiplying $\vec{G}$ by $-4$ inside the integral is equivalent to multiplying the integral of $\vec{G}$ by $-4$.

If F⃗=−4G⃗\vec{F}=-4 \vec{G}F=−4G And S Is an Oriented Surface, Then ∫SF⃗⋅dA⃗=−4∫SG⃗⋅dA⃗\int _ { S } \vec { F } \cdot \vec { d A } = - 4 \int _ { S } \vec { G } \cdot \vec { d A }∫S​F⋅dA=−4∫S​G⋅dA

If $\vec{F}=-4 \vec{G}$ And S Is an Oriented Surface, Then $\int _ { S } \vec { F } \cdot \vec { d A } = - 4 \int _ { S } \vec { G } \cdot \vec { d A }$