The area bounded by the coordinate axes and the curve $\sqrt {\rm{x}} + \sqrt {\rm{y}} = 1$ , is

Question

Prepp Team · Accepted Answer

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:

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

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

Revision Table: Area Under Curve

Concept	Description	Formula/Method
Area Bounded by Curve & x-axis	Area between $y = f(x)$, x-axis, $x=a$, and $x=b$.	$\int_{a}^{b} \|f(x)\| \, dx$
Area Bounded by Curve & y-axis	Area between $x = g(y)$, y-axis, $y=c$, and $y=d$.	$\int_{c}^{d} \|g(y)\| \, dy$
Integration Limits	Determined by intersection points or specified boundaries.	Find x-intercepts for dx integral, y-intercepts for dy integral.

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).

Prepp Team · Answer

1 square unit

Prepp Team · Answer

$\frac{1}{2}$ square unit

Prepp Team · Answer

$\frac{1}{3}$ square unit