The domain of the vector field $\mathbf { F } ( x , y ) = \sin ( x + y ) \mathbf { i } - \frac { 5 } { x ^ { 2 } - 2 x y + y ^ { 2 } } \mathbf { j }$ is

A) $\{ ( x , y ) : x \neq y \}$
B) $\{ ( x , y ) : x y 0 \}$
C) $\{ ( x , y ) : x \neq | y | \}$
D) $\{ ( x , y ) : x 1 , y 0 \}$
E) $\{ ( x , y ) : x \geq | y | \}$

Question

The domain of the vector field $\mathbf { F } ( x , y ) = \sin ( x + y ) \mathbf { i } - \frac { 5 } { x ^ { 2 } - 2 x y + y ^ { 2 } } \mathbf { j }$ is

A) $\{ ( x , y ) : x \neq y \}$
B) $\{ ( x , y ) : x y > 0 \}$
C) $\{ ( x , y ) : x \neq | y | \}$
D) $\{ ( x , y ) : x > 1 , y > 0 \}$
E) $\{ ( x , y ) : x \geq | y | \}$

Quizplus · Accepted Answer

The domain is determined by where the vector field is defined. The only restriction comes from the denominator of the second component, $x^2 - 2xy + y^2$, which must not be zero. This expression can be rewritten as $(x - y)^2$, which is zero when $x = y$. Therefore, the domain is all $(x, y)$ such that $x \neq y$.

The Domain of the Vector Field F(x,y)=sin⁡(x+y)i−5x2−2xy+y2j\mathbf { F } ( x , y ) = \sin ( x + y ) \mathbf { i } - \frac { 5 } { x ^ { 2 } - 2 x y + y ^ { 2 } } \mathbf { j }F(x,y)=sin(x+y)i−x2−2xy+y25​j

The Domain of the Vector Field $\mathbf { F } ( x , y ) = \sin ( x + y ) \mathbf { i } - \frac { 5 } { x ^ { 2 } - 2 x y + y ^ { 2 } } \mathbf { j }$