Example 1 | Plane Areas in Rectangular Coordinates

Tags:

horizontal strip

limits of integration

downward parabola

parabolic segment

Example 1
Find the area bounded by the curve y = 9 - x² and the x-axis.

Solution

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Step 1: Sketch the curve.

$y = 9 - x^2$

$x^2 = -y + 9$

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

Area bounded by downward parabola and x-axis

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.

The strip shown will start from y = 0 and end to y = 9

Step 3: Apply the appropriate formula then integrate.

$A = \displaystyle {\int_{y_1}}^{y_2} (x_R - x_L) \, dy$

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

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Step 2: Determine the limits of the strip.

In this case the limits are not defined; we need to solve the points of intersection of the curves.
$y = 9 - x^2$

when y = 0, x = &pm; 3. The strip will swipe from x = 0 to x = 3.

Step 3: Apply the appropriate formula then integrate.

$A = {\int_{x_1}}^{x_2} (y_U - y_L) \, dx$

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!

Submitted by Romel Verterra on February 22, 2010 - 11:54pm

20685 reads

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