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Chapter 13: Problem 25

Describe the region \(R\) in the \(x y\) -plane that corresponds to the domain of the function. $$ h(x, y)=x \sqrt{y} $$

Short Answer

Expert verified
The domain of the function \(h(x, y)=x \sqrt{y}\) corresponds to the region \(R\) in the \(x y\)-plane where \(x\) can be any real number and \(y\) is a non-negative real number.

Step by step solution

01

Understanding Defined Regions

The domain of a multi-variable function is the set of all ordered pairs \((x, y)\) such that the function is defined. For the function \(h(x, y)=x \sqrt{y}\), the domain will be the ordered pairs \((x, y)\) for which the square root of \(y\) is defined.
02

Define Condition for Square Roots

In mathematics, the square root of a number is only defined for non-negative numbers. Hence, \(y\) should be greater than or equal to zero. Also, \(x\) is not under the root, thus its values can be any real number.
03

Define the Domain

Thus, the domain \((x, y)\) of the function \(h(x, y)=x \sqrt{y}\) is defined by the set of all ordered pairs \((x, y)\) where \(x\) is any real number and \(y\) is a non-negative real number.

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Most popular questions from this chapter

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{-2}^{2} \int_{0}^{4-y^{2}} d x d y $$

Evaluate the partial integral. $$ \int_{x}^{x^{2}} \frac{y}{x} d y $$

Set up the integral for both orders of integration and use the more convenient order to evaluate the integral over the region \(R\). $$ \begin{aligned} &\int_{R} \int \frac{y}{1+x^{2}} d A\\\ &R: \text { region bounded by } y=0, y=\sqrt{x}, x=4 \end{aligned} $$

Find the average value of \(f(x, y)\) over the region \(R\). $$ \begin{aligned} &f(x, y)=x y\\\ &R: \text { rectangle with vertices }(0,0),(4,0),(4,2),(0,2) \end{aligned} $$

Sketch the region of integration and evaluate the double integral. $$ \int_{-a}^{a} \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2-x^{2}}}} d y d x $$
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