Find the area enclosed by the circle x^{2}+y^ | vedclass

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(A) The given equation of the circle is $x^{2}+y^{2}=a^{2}$.Since the circle is symmetrical about both the $x$-axis and the $y$-axis,the total area is

Vedclass · Accepted Answer

$\pi a^{2}$

Vedclass · Answer

(A) The given equation of the circle is $x^{2}+y^{2}=a^{2}$.Since the circle is symmetrical about both the $x$-axis and the $y$-axis,the total area is $4$ times the area of the region $AOBA$ in the first quadrant.Area $= 4 \int_{0}^{a} y \, dx$Since $x^{2}+y^{2}=a^{2}$,we have $y = \sqrt{a^{2}-x^{2}}$ in the first quadrant.Area $= 4 \int_{0}^{a} \sqrt{a^{2}-x^{2}} \, dx$Using the standard integral formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right)$:Area $= 4 \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right) \right]_{0}^{a}$$= 4 \left[ \left( \frac{a}{2} \sqrt{a^{2}-a^{2}} + \frac{a^{2}}{2} \sin^{-1}(1) \right) - (0 + 0) \right]$$= 4 \left[ 0 + \frac{a^{2}}{2} \left(\frac{\pi}{2}\right) \right]$$= 4 \left( \frac{\pi a^{2}}{4} \right) = \pi a^{2}$Thus,the area enclosed by the circle is $\pi a^{2}$ square units.

Find the area enclosed by the circle $x^{2}+y^{2}=a^{2}$.

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