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In Exercises \(31-36,\) use an iterated integral to find the area of the region bounded by the graphs of the equations. $$ \sqrt{x}+\sqrt{y}=2, \quad x=0, \quad y=0 $$

Short Answer

Expert verified
The area of the region bounded by the given equations is \( \frac{16}{3} \) square units.

Step by step solution

01

Isolate y

Rearrange the given equation \( \sqrt{x} + \sqrt{y} = 2 \) to isolate y. This helps identify the integral boundaries for y. As a result, the equation becomes \( y = (2 - \sqrt{x})^2 \) or \( y = 4 - 4\sqrt{x} + x \).
02

Set Up the Iterated Integral

Now, set up the iterated integral for the given region using the newly created equation. With a double integral, the integrand is 1, as the task is to find the area. The limits of x are from 0 to 4 (from the y-equation where \( y = 0 \)), and the limits of y are from 0 to \(4 - 4\sqrt{x} + x\). The entire integral looks like this: \[ \int_0^4 \int_0^{4 -4\sqrt{x} + x} 1 \, dy \, dx \].
03

Evaluate the Integral

Evaluate the inner integral (y-integral) first by applying the fundamental theorem of calculus. Then, evaluate the outer integral (x-integral). After evaluating these integrals, we have: \[ \frac{16}{3} \].

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