Area of the region in the first quadrant enclosed by the X-axis, the line y = x and the circle x2 + y2 = 32 is:

Question

Area of the region in the first quadrant enclosed by the $X$ -axis, the line $y = x$ and the circle $x^{2} + y^{2} = 32$ is:

Smart Achievers · Accepted Answer

(2)

Ans. $4 π$ sq. units

Explanation:

We have, area enclosed by $X$ -axis i.e., $y = 0$ , $y = x$ and the circle $x^{2} + y^{2} = 32$ in first quadrant.

Since, $x^{2} + {(x)}^{2} = 32$ $[∵ y = x]$

$\Rightarrow$ $2 x^{2} = 32$

$\Rightarrow$ $x = \pm 4$

So, the intersection point of circle $x^{2} + y^{2} = 32$ and line $y = x$ are (4, 4) or $(- 4, - 4)$ .

and $x^{2} + y^{2} = {(4 \sqrt{2})}^{2}$

Since, $y = 0$

$∴$ $x^{2} + {(0)}^{2} = 32$

$\Rightarrow$ $x = \pm 4 \sqrt{2}$

So, the circle intersects the $X$ -axis at $(\pm 4 \sqrt{2}, 0)$ .

Area of shaded region

$= \int_{0}^{4} x d x + \int_{4}^{4 \sqrt{2}} \sqrt{{(4 \sqrt{2})}^{2} - x^{2}} d x$

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

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

$= 8 + [16 \cdot \frac{π}{2} - 2 \sqrt{16} - 16 \cdot \frac{π}{4}]$

$= 8 + [8 π - 8 - 4 π] = 4 π sq. units.$

Solution

Q.
Area of the region in the first quadrant enclosed by the $X$ -axis, the line $y = x$ and the circle $x^{2} + y^{2} = 32$ is:

1 $16 π$ sq. units

2 $4 π$ sq. units

3 $32 π$ sq. units

4 $24 π$ sq. units

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Solution

Q. Area of the region in the first quadrant enclosed by the X-axis, the line y=x and the circle x2+y2=32 is:

1 16π sq. units

2 4π sq. units

3 32π sq. units