Question

Answer the question below​

Answer 1

Answer:

Refer the attachement for better understanding

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Be Brainly!

Answer 2

EXPLANATION.

Area bounded by x² + y² = 4.

x = √3y and x-axis.

The area should be in the first quadrant.

As we know that,

Equation : x² + y² = 4.

This is the equation of circle.

⇒ x² + y² = (2)².

Radius of circle = 2.

Put the value of x = √3y in equation, we get.

⇒ (√3y)² + y² = 4.

⇒ 3y² + y² = 4.

⇒ 4y² = 4.

⇒ y² = 1.

⇒ y = ± 1.

Put the value of y = 1 in equation, we get.

⇒ x = √3y.

⇒ x = √3(1).

⇒ x = √3.

Put the value of y = -1 in equation, we get.

⇒ x = √3y.

⇒ x = √3(-1).

⇒ x = -√3.

⇒ x = √3y.

⇒ y = x/√3.

⇒ x² + y² = 4.

⇒ y² = 4 - x².

⇒ y = ±√4 - x².

For equation considered as 1st quadrant then y = √4 - x².

⇒ A = A₁ + A₂.

Area of OXY = Area of OXZ - Area of ZXY.

$\sf \implies \displaystyle \int_{0}^{\sqrt{3} } y dx = Area \ OXZ.$

$\sf \implies \displaystyle \int_{0}^{\sqrt{3} } \dfrac{x}{\sqrt{3} } dx.$

$\sf \implies \dfrac{1}{\sqrt{3} } \displaystyle \int_{0}^{\sqrt{3} } x dx$

$\sf \implies \dfrac{1}{\sqrt{3} } \bigg[\dfrac{x^{2} }{2} \bigg]_{0}^{\sqrt{3}}$

$\sf \implies \displaystyle \dfrac{1}{2\sqrt{3} } \bigg[ x^{2} \bigg]_{0}^{\sqrt{3} }$

$\sf \implies \displaystyle \dfrac{1}{2\sqrt{3} } \bigg[(\sqrt{3} )^{2} - (0)^{2} \bigg]$

$\sf \implies \displaystyle \dfrac{1}{2\sqrt{3} } \times 3 = \dfrac{\sqrt{3}}{2}$

$\sf \implies \displaystyle \int_{\sqrt{3} }^{2} ydx. \ = Area \ of \ ZXY.$

$\sf \implies \displaystyle \int_{\sqrt{3} }^{2} \sqrt{4 - x^{2} } dx$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \sqrt{a^{2} - x^{2} } dx = \dfrac{1}{2}x\sqrt{a^{2} - x^{2} } +\dfrac{a^{2} }{2} sin^{-1} \bigg(\dfrac{x}{a} \bigg) + C.$

Using this formula in equation, we get.

$\sf \implies \displaystyle \int_{\sqrt{3} }^{2} \sqrt{(2)^{2} - x^{2} } dx$

$\sf \implies \displaystyle \bigg[ \dfrac{1}{2} x \sqrt{4 - x^{2} } + 2 sin^{-1} \bigg(\frac{x}{2} \bigg) \bigg]_{\sqrt{3} }^{2}$

$\sf \implies \displaystyle \bigg[\dfrac{1}{2} (2 ) \sqrt{4 - (2)^{2} } + 2 sin^{-1} \bigg(\dfrac{2}{2} \bigg) \bigg] \ - \bigg[ \dfrac{1}{2} (\sqrt{3} ) \sqrt{4 - (\sqrt{3})^{2} } + 2sin^{-1} \bigg(\dfrac{\sqrt{3} }{2} \bigg) \bigg]$

$\sf \implies \displaystyle \bigg[ 0 + 2sin^{-1} (1) \bigg] \ - \bigg[ \dfrac{\sqrt{3} }{2} + 2sin^{-1}\bigg(\dfrac{\sqrt{3} }{2} \bigg) \bigg]$

$\sf \implies \displaystyle \bigg[ 2 sin^{-1} (1) - \dfrac{\sqrt{3} }{2} - 2sin^{-1} \bigg(\dfrac{\sqrt{3} }{2} \bigg) \bigg].$

$\sf \implies \displaystyle - \dfrac{\sqrt{3} }{2} + 2 \bigg[ sin^{-1} (1) - sin^{-1} \bigg( \dfrac{\sqrt{3} }{2} \bigg) \bigg]$

$\sf \implies \displaystyle \dfrac{-\sqrt{3} }{2} + 2 \bigg[ \dfrac{\pi}{2} - \dfrac{\pi}{3} \bigg]$

$\sf \implies \displaystyle \dfrac{-\sqrt{3} }{2} + 2 \bigg[ \dfrac{\pi}{6} \bigg] \ = \dfrac{-\sqrt{3} }{2} + \dfrac{\pi}{3}$

Area of OYX = Area of OXZ - Area of ZXY.

$\sf \implies \displaystyle \dfrac{\sqrt{3} }{2} - \frac{\sqrt{3} }{2} + \dfrac{\pi}{3} = \dfrac{\pi}{3} \ sq. units.$