Name: Area Under Curve: Illustration
Uploaded: 2019-07-19
Duration: 3 min 12 s
Description: We all know that the area of a circle of radius r is given by the formula π(r)^2. This informative video explains how to get the area under a curve using def...

Question

Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line

y = x

, and the circle

x^{2} + y^{2} = 32

Accepted Answer

Given the boundaries of the area to be found are,

the circle, $x^{2} + y^{2} = 32$

the line, $y = x . . . . . . . . . (2)$

the $x$ -axis

And, area should be in first quadrant.

From the equation, of the first circle, $x^{2} + y^{2} = 32$

the vertex at $(0, 0)$ i.e. the origin

the radius is $4 \sqrt{2} unit$

Now to find the point of intersection of $(1)$ and $(2)$ substitute $y = x$ in $(1)$

$x^{2} + x^{2} = 32$

$2 x^{2} = 32$

$x^{2} = 16$

$x = \pm 4$

Substituting $x$ in $(2),$ we get $y = \pm 4$

So the two points, $A$ and $B$ where $(1)$ and $(2)$ meet are $A = (4, 4)$ and $B = (- 4, - 4)$

As $x$ and $y$ have even powers for both the circles, they will be symmetrical about the $x -$ axis and $y -$ axis.

Consider the circle, $x^{2} + y^{2} = 32$ can be re-written as

$y^{2} = 32 - x^{2}$

$y = \sqrt{(32 - x^{2})}$

Let us drop a perpendicular from $A$ on to $x -$ axis. The base of the perpendicular is $C = (4, 0)$

Area of the required region $=$ Area of $O A D O .$

Area of $O A D O =$ Area of $C A D C +$ Area of $O A C$

Area of $OADO = \int_{4}^{4 \sqrt{2}} \sqrt{32 - x^{2}} dx + \int_{0}^{4} xdx$

$= {[\frac{x \sqrt{32 - x^{2}}}{2} + \frac{32}{2} \sin^{- 1} (\frac{x}{4 \sqrt{2}})]}_{4}^{4 \sqrt{2}} + {[\frac{x^{2}}{2}]}_{0}^{4}$

Using the formula, $\int \sqrt{a^{2} - x^{2}} dx = \frac{x \sqrt{a^{2} - x^{2}}}{2} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a}) + c$ and $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$

$=$ $[\frac{4 \sqrt{2} \sqrt{32 - 32}}{2} + 16 \sin^{- 1} (\frac{4 \sqrt{2}}{4 \sqrt{2}}) - (\frac{4 \sqrt{32 - 16}}{2} + 16 \sin^{- 1} (\frac{1}{\sqrt{2}}))] + [\frac{16}{2} - 0]$

$= [0 + 16 \times \frac{π}{2} - (8 + 16 \times \frac{π}{4})] + 8$

$= 4 π$

Hence, area of the required region is $4 π$ square units.

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