Finding Area Bounded by Curve and Axes

The problem asks us to find the area bounded by the curve given by the equation \(\sqrt {\rm{x}} + \sqrt {\rm{y}} = 1\) and the coordinate axes. The coordinate axes are the x-axis (\(\rm{y}=0\)) and the y-axis (\(\rm{x}=0\)). The area bounded by the curve and the positive coordinate axes is typically in the first quadrant.

Step 1: Determine the Intercepts with the Axes

To find the points where the curve intersects the coordinate axes, we set \(\rm{x}=0\) and \(\rm{y}=0\) in the equation of the curve.

• Intersection with the y-axis (when \(\rm{x}=0\)):
  \(\sqrt{0} + \sqrt{\rm{y}} = 1\)
  \(\sqrt{\rm{y}} = 1\)
  Squaring both sides, \(\rm{y} = 1^2 = 1\).
  The curve intersects the y-axis at the point (0, 1).
• Intersection with the x-axis (when \(\rm{y}=0\)):
  \(\sqrt{\rm{x}} + \sqrt{0} = 1\)
  \(\sqrt{\rm{x}} = 1\)
  Squaring both sides, \(\rm{x} = 1^2 = 1\).
  The curve intersects the x-axis at the point (1, 0).

So, the area we need to find is bounded by the curve \(\sqrt {\rm{x}} + \sqrt {\rm{y}} = 1\), the x-axis (from \(\rm{x}=0\) to \(\rm{x}=1\)), and the y-axis (from \(\rm{y}=0\) to \(\rm{y}=1\)).

Step 2: Express y in terms of x

To find the area using integration with respect to x, we need to express \(\rm{y}\) as a function of \(\rm{x}\) from the curve's equation \(\sqrt {\rm{x}} + \sqrt {\rm{y}} = 1\).

\(\sqrt {\rm{y}} = 1 - \sqrt {\rm{x}}\)

Squaring both sides to solve for \(\rm{y}\):

\(\rm{y} = (1 - \sqrt {\rm{x}})^2\)

Expand the right side:

\(\rm{y} = (1)^2 - 2(1)(\sqrt{\rm{x}}) + (\sqrt{\rm{x}})^2\)

\(\rm{y} = 1 - 2\sqrt{\rm{x}} + \rm{x}\)

We can write \(\sqrt{\rm{x}}\) as \(\rm{x}^{1/2}\):

\(\rm{y} = 1 - 2\rm{x}^{1/2} + \rm{x}\)

Step 3: Set up the Definite Integral for Area Calculation

The area \(\rm{A}\) bounded by the curve \(\rm{y} = f(\rm{x})\), the x-axis, and the vertical lines \(\rm{x} = \rm{a}\) and \(\rm{x} = \rm{b}\) is given by the definite integral \(\int_{\rm{a}}^{\rm{b}} \rm{y} \, d\rm{x}\). In our case, the curve is \(\rm{y} = 1 - 2\rm{x}^{1/2} + \rm{x}\), the lower limit is \(\rm{x}=0\) (y-axis), and the upper limit is \(\rm{x}=1\) (x-intercept).

So, the area is:

\(\rm{A} = \int_0^1 (1 - 2\rm{x}^{1/2} + \rm{x}) \, d\rm{x}\)

Step 4: Evaluate the Definite Integral

Now, we evaluate the integral:

\(\rm{A} = \int_0^1 (1 - 2\rm{x}^{1/2} + \rm{x}) \, d\rm{x}\)

Integrate term by term:

• The integral of 1 is \(\rm{x}\).
• The integral of \(-2\rm{x}^{1/2}\) is \(-2 \frac{\rm{x}^{1/2+1}}{1/2+1} = -2 \frac{\rm{x}^{3/2}}{3/2} = -2 \times \frac{2}{3} \rm{x}^{3/2} = -\frac{4}{3}\rm{x}^{3/2}\).
• The integral of \(\rm{x}\) is \(\frac{\rm{x}^2}{2}\).

So, the indefinite integral is \(\rm{x} - \frac{4}{3}\rm{x}^{3/2} + \frac{\rm{x}^2}{2}\).

Now, evaluate the definite integral from 0 to 1:

\(\rm{A} = \left[ \rm{x} - \frac{4}{3}\rm{x}^{3/2} + \frac{\rm{x}^2}{2} \right]_0^1\)

Substitute the upper limit (\(\rm{x}=1\)) and subtract the result of substituting the lower limit (\(\rm{x}=0\)):

\(\rm{A} = \left( 1 - \frac{4}{3}(1)^{3/2} + \frac{(1)^2}{2} \right) - \left( 0 - \frac{4}{3}(0)^{3/2} + \frac{(0)^2}{2} \right)\)

\(\rm{A} = \left( 1 - \frac{4}{3}(1) + \frac{1}{2} \right) - (0 - 0 + 0)\)

\(\rm{A} = \left( 1 - \frac{4}{3} + \frac{1}{2} \right) - 0\)

Find a common denominator for the fractions (which is 6):

\(\rm{A} = \left( \frac{6}{6} - \frac{8}{6} + \frac{3}{6} \right)\)

\(\rm{A} = \frac{6 - 8 + 3}{6}\)

\(\rm{A} = \frac{1}{6}\)

The area bounded by the curve \(\sqrt {\rm{x}} + \sqrt {\rm{y}} = 1\) and the coordinate axes is \(\frac{1}{6}\) square unit.

Summary of Area Calculation Steps

Here is a summary of the steps taken to find the area:

1. Identified the boundaries: the curve, x-axis, and y-axis.
2. Found the intercepts of the curve with the axes to determine the limits of integration.
3. Rearranged the curve equation to express y as a function of x.
4. Set up the definite integral \(\int \rm{y} \, d\rm{x}\) with the appropriate limits.
5. Evaluated the definite integral to find the area.

The calculated area is \(\frac{1}{6}\) square unit.

Revision Table: Area Under Curve

Additional Information: The Curve \(\sqrt{\rm{x}} + \sqrt{\rm{y}} = 1\)

The equation \(\sqrt{\rm{x}} + \sqrt{\rm{y}} = 1\) represents a part of a parabola rotated in the first quadrant. If we square both sides repeatedly, we get:

\(\sqrt{\rm{y}} = 1 - \sqrt{\rm{x}}\)

\(\rm{y} = (1 - \sqrt{\rm{x}})^2 = 1 - 2\sqrt{\rm{x}} + \rm{x}\)

Rearranging terms to isolate the square root:

\(2\sqrt{\rm{x}} = 1 + \rm{x} - \rm{y}\)

Squaring again:

\((2\sqrt{\rm{x}})^2 = (1 + \rm{x} - \rm{y})^2\)

\(4\rm{x} = (1 + \rm{x})^2 - 2(1 + \rm{x})\rm{y} + \rm{y}^2\)

\(4\rm{x} = 1 + 2\rm{x} + \rm{x}^2 - 2\rm{y} - 2\rm{xy} + \rm{y}^2\)

\(\rm{x}^2 - 2\rm{xy} + \rm{y}^2 - 2\rm{x} - 2\rm{y} + 1 = 0\)

This is a quadratic form in x and y, which represents a conic section. The specific form suggests it's a parabola rotated by 45 degrees. The region \(\sqrt{\rm{x}} + \sqrt{\rm{y}} = 1\) for \(\rm{x} \ge 0\) and \(\rm{y} \ge 0\) is the arc connecting (1,0) and (0,1).