The domain of the vector field $\mathbf { F } ( x , y ) = \sqrt { 1 - x ^ { 2 } } \mathbf { i } + \ln y \mathbf { j }$ is

A) $\{ ( x , y ) : x \langle 1 , y \rangle 0 \}$
B) $\{ ( x , y ) : x y 0 \}$
C) $\{ ( x , y ) : - 1 \leq x \leq 1 , y 0 \}$
D) $\{ ( x , y ) : x \leq 1 , y \geq 0 \}$
E) $\{ ( x , y ) : - 1 \leq x \langle 1 , y \rangle 0 \}$

Question

The domain of the vector field $\mathbf { F } ( x , y ) = \sqrt { 1 - x ^ { 2 } } \mathbf { i } + \ln y \mathbf { j }$ is

A) $\{ ( x , y ) : x \langle 1 , y \rangle 0 \}$
B) $\{ ( x , y ) : x y > 0 \}$
C) $\{ ( x , y ) : - 1 \leq x \leq 1 , y > 0 \}$
D) $\{ ( x , y ) : x \leq 1 , y \geq 0 \}$
E) $\{ ( x , y ) : - 1 \leq x \langle 1 , y \rangle 0 \}$

Quizplus · Accepted Answer

The domain is determined by the conditions under which the expressions inside the square root and the logarithm are defined. For $\sqrt{1-x^2}$, $1-x^2$ must be non-negative, which means $-1 \leq x \leq 1$. For $\ln y$, $y$ must be positive, so $y > 0$. Therefore, the correct domain is $\{ ( x , y ) : - 1 \leq x \leq 1 , y > 0 \}$.

The Domain of the Vector Field F(x,y)=1−x2i+ln⁡yj\mathbf { F } ( x , y ) = \sqrt { 1 - x ^ { 2 } } \mathbf { i } + \ln y \mathbf { j }F(x,y)=1−x2​i+lnyj

The Domain of the Vector Field $\mathbf { F } ( x , y ) = \sqrt { 1 - x ^ { 2 } } \mathbf { i } + \ln y \mathbf { j }$