Find the area of the region in the first quad | vedclass

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(A) The given equations are:$y=x$  ........... $(1)$$x^{2}+y^{2}=32$    ........... $(2)$Solving $(1)$ and $(2)$,we substitute $y=x$ in $(2)$ to get $

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$4$

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(A) The given equations are:$y=x$  ........... $(1)$$x^{2}+y^{2}=32$    ........... $(2)$Solving $(1)$ and $(2)$,we substitute $y=x$ in $(2)$ to get $x^{2}+x^{2}=32$,which gives $2x^{2}=32$,so $x^{2}=16$. Since we are in the first quadrant,$x=4$. Thus,the line and the circle meet at $B(4, 4)$.Draw perpendicular $BM$ to the $x-$ axis,where $M$ is $(4, 0)$.The required area is the area of the region $OBMO$ $+$ area of the region $BMAB$.Area of region $OBMO = \int_{0}^{4} y \, dx = \int_{0}^{4} x \, dx = \left[\frac{x^{2}}{2}ight]_{0}^{4} = \frac{16}{2} = 8$.Area of region $BMAB = \int_{4}^{4\sqrt{2}} y \, dx = \int_{4}^{4\sqrt{2}} \sqrt{(4\sqrt{2})^{2}-x^{2}} \, dx$.Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}ight)$:$= \left[\frac{x}{2}\sqrt{32-x^{2}} + \frac{32}{2}\sin^{-1}\left(\frac{x}{4\sqrt{2}}ight)ight]_{4}^{4\sqrt{2}}$$= \left(0 + 16 \sin^{-1}(1)ight) - \left(\frac{4}{2}\sqrt{32-16} + 16 \sin^{-1}\left(\frac{4}{4\sqrt{2}}ight)ight)$$= 16 	imes \frac{\pi}{2} - \left(2 	imes 4 + 16 	imes \frac{\pi}{4}ight) = 8\pi - (8 + 4\pi) = 4\pi - 8$.Total area $= 8 + (4\pi - 8) = 4\pi$.

Find the area of the region in the first quadrant enclosed by the $x-$ axis,the line $y=x,$ and the circle $x^{2}+y^{2}=32$. (in $\pi$)

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