Find the domain and sketch for $z=x+\sqrt{y}$ and $z=\arccos y$

Expert intro: Domain questions in multivariable settings are mostly about reading the restrictions carefully: square roots demand nonnegative inputs, and inverse trig functions come with their own interval constraints.

Detailed walkthrough

We check each function separately.

(a) $z=x+\sqrt{y}$

The term $x$ is unrestricted, but $\sqrt{y}$ requires

$y\ge 0.$

So the domain is

$\boxed{\{(x,y)\in\mathbb R^2: y\ge 0\}}.$

Sketch description

• All points in the half-plane on or above the $x$-axis.
• The boundary line $y=0$ is included.

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(b) $z=\arccos y$

For the inverse cosine to be defined,

$-1\le y\le 1.$

There is no restriction on $x$ because $x$ does not appear in the formula. So the domain is

$\boxed{\{(x,y)in\mathbb R^2: -1\le y\le 1\}}.$

Sketch description

• A horizontal strip between the lines $y=-1$ and $y=1$.
• Both boundary lines are included.

If you are drawing these on the $xy$-plane, it helps to shade the allowed region rather than try to graph $z$ itself.

💡 Pitfall guide

A frequent mistake is to treat $z=\arccos y$ as though only the variable $y$ matters and forget that the domain lives in the $xy$-plane. Another one: excluding the boundary for $\sqrt{y}$ or for $\arccos y$. Both endpoints belong.

🔄 Real-world variant

If the first function were $z=x+\sqrt{y-2}$, the domain would shift to $y\ge 2$. If the second were $z=\arccos(2y)$, then the condition becomes $-1\le 2y\le 1$, so $-\tfrac12\le y\le \tfrac12$.

🔍 Related terms

domain of definition, square root restriction, inverse cosine domain