# Example 1 | Plane Areas in Rectangular Coordinates

**Example 1**

Find the area bounded by the curve y = 9 - x^{2} and the x-axis.

**Solution**

$x^2 = -y + 9$

$x^2 = -(y - 9)$ → downward parabola; vertex at (0, 9); latus rectum = 1

The required area is symmetrical with respect to the y-axis, in this case, integrate the half of the area then double the result to get the total area. The use of symmetry will greatly simplify our solution most especially to curves in **polar coordinates**.

**Using Horizontal Strip**

Step 2: Determine the limits of the strip.

Step 3: Apply the appropriate formula then integrate.

Where

$y_1 = 0$

$y_2 = 9$

$x_R = \text{parabola} = (9 - y)^{1/2}$

$x_L = \text{y-axis} = 0$

$A = \displaystyle 2{\int_{0}}^{\,9} [ \, (9 - y)^{1/2} - 0 \, ] \, dy$

$A = \displaystyle 2{\int_{0}}^{\,9} (9 - y)^{1/2} \, dy$

$A = \displaystyle -2{\int_{0}}^{\,9} (9 - y)^{1/2} \, (-dy)$

$A = -2\left[ \dfrac{(9 - y)^{3/2}}{3/2} \right]_0^9$

$A = -\frac{4}{3} [ \, (9 - 9)^{3/2} - (9 - 0)^{3/2} \, ]$

$A = 36 \, \text{ unit}^2$ *answer*

**Using Vertical Strip**

$y = 9 - x^2$

when y = 0, x = ± 3. The strip will swipe from x = 0 to x = 3.

Step 3: Apply the appropriate formula then integrate.

Where

$x_1 = 0$

$x_2 = 3$

$y_U = \text{parabola} = 9 - x^2$

$y_L = \text{x-axis} = 0$

$A = \displaystyle 2 {\int_{0}}^{\,3} [ \, (9 - x^2) - 0 \, ] \, dx$

$A = \displaystyle 2 {\int_{0}}^{\,3} (9 - x^2) \, dx$

$A = 2 \left[ 9x - \dfrac{x^3}{3} \right]_0^3$

$A = 2 \bigg\{ \left[ 9(3) - \dfrac{3^3}{3} \right] - \left[ 9(0) - \dfrac{0^3}{3} \right] \bigg\}$

$A = 36 \, \text{unit}^2$ *okay!*