Find the areas of the regions enclosed by the curves and lines in Exercises 15–26.
√x + √y = 1, x = 0, y = 0

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Rewrite the given curve equation \(\sqrt{x} + \sqrt{y} = 1\) to express \(y\) in terms of \(x\). Start by isolating \(\sqrt{y}\): \(\sqrt{y} = 1 - \sqrt{x}\).

Square both sides to solve for \(y\): \(y = (1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + x\).

Identify the region bounded by the curve and the coordinate axes: \(x=0\), \(y=0\), and the curve \(y = (1 - \sqrt{x})^2\) from \(x=0\) to \(x=1\).

Set up the integral for the area of the region as \(\int_0^1 y \, dx = \int_0^1 (1 - 2\sqrt{x} + x) \, dx\).

Evaluate the integral step-by-step (without calculating the final value): integrate each term separately: \(\int_0^1 1 \, dx\), \(\int_0^1 -2\sqrt{x} \, dx\), and \(\int_0^1 x \, dx\), then sum the results to find the total area.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Area Between Curves

The area between curves is found by integrating the difference between the upper and lower functions over a given interval. When bounded by axes and a curve, the area can be computed by setting up an integral with appropriate limits reflecting the region.

Implicit Functions and Curve Manipulation

The given curve √x + √y = 1 is implicit. To find the area, it is often necessary to express y as a function of x or vice versa. This involves isolating one variable and squaring both sides carefully to avoid extraneous solutions.

Definite Integration with Variable Substitution

Calculating the area under curves involving roots often requires substitution to simplify the integral. For example, substituting u = √x or v = √y can transform the integral into a more manageable form, facilitating evaluation.

Find the areas of the regions enclosed by the curves and lines in Exercises 15–26.√x + √y = 1, x = 0, y = 0