using integrals find the area of the region { (x,y): 9x^2 +y^2<=36 and 3x+y>=6} urgent please

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ 9{\mathrm{x}}^2+{\mathrm{y}}^2\le 36 \\ \Rightarrow \frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{36}\le 1 \\ \Rightarrow \frac{{\mathrm{x}}^2}{2^2}+\frac{{\mathrm{y}}^2}{6^2}\le 1 \\ \mathrm{Ellipse} \mathrm{will} \mathrm{be} \mathrm{verical} \mathrm{as} \mathrm{b}>\mathrm{a} \\ 3\mathrm{x}+\mathrm{y}\ge 6 \\ \Rightarrow \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{6}\ge 1 \\ \mathrm{Equation} \mathrm{intercepts} \mathrm{X}\hspace{0.17em}\mathrm{and} \mathrm{Y}\hspace{0.17em}\mathrm{axis} \mathrm{at} \left(2,0\right) \mathrm{and} \left(0,6\right) \mathrm{where} \mathrm{it} \mathrm{will} \mathrm{ellipse} \mathrm{too} \\ \left[\mathrm{We} \mathrm{can} \mathrm{easily} \mathrm{see} \mathrm{it} \mathrm{by} \mathrm{drawing} \mathrm{graph}.\right] \\ \mathrm{So} \mathrm{we} \mathrm{now} \mathrm{know} \mathrm{which} \mathrm{are} \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{find}. \\ \mathrm{Consider} \\ 9{\mathrm{x}}^2+{\mathrm{y}}^2=36 \\ \Rightarrow {\mathrm{y}}^2=36-9{\mathrm{x}}^2=9\left(4-{\mathrm{x}}^2\right) \\ \Rightarrow \mathrm{y}=\pm 3\sqrt{4-{\mathrm{x}}^2} \\ \mathrm{But} \mathrm{we} \mathrm{are} \mathrm{concerned} \mathrm{about} \mathrm{first} \mathrm{quadrant} \mathrm{only}: \\ {\mathrm{y}}_1=3\sqrt{4-{\mathrm{x}}^2} \\ 3\mathrm{x}+\mathrm{y}=6 \\ \Rightarrow {\mathrm{y}}_2=6-3\mathrm{x}=3\left(2-\mathrm{x}\right) \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{region}={\int }_0^2\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\mathrm{dx} \\ ={\int }_0^2\left[3\sqrt{4-{\mathrm{x}}^2}-3\left(2-\mathrm{x}\right)\right]\mathrm{dx} \\ =3{\int }_0^2\left[\sqrt{2^2-{\mathrm{x}}^2}-\left(2-\mathrm{x}\right)\right]\mathrm{dx} \\ =3{\int }_0^2\left[\sqrt{2^2-{\mathrm{x}}^2}+\left(\mathrm{x}-2\right)\right]\mathrm{dx} \\ =3{\left[\frac{\mathrm{x}}{2}\sqrt{2^2-{\mathrm{x}}^2}+\frac{2^2}{2}{\sin }^{-1}\left(\frac{\mathrm{x}}{2}\right)+\frac{{\left(\mathrm{x}-2\right)}^2}{2}\right]}_0^2 \\ =3\left[\left\{\frac{2}{2}\sqrt{4-4}+2{\sin }^{-1}\left(1\right)+\frac{{\left(2-2\right)}^2}{2}\right\}-\left\{0+0+\frac{{\left(2-0\right)}^2}{2}\right\}\right] \\ =3\left[\left\{0+2.\frac{\mathrm{\pi }}{2}-0\right\}-\left\{2\right\}\right] \\ \mathbf{=}\mathbf{3}\left[\mathbf{\pi }\mathbf{-}\mathbf{2}\right] \\ \mathbf{=}\mathbf{3}\mathbf{\pi }\mathbf{-}\mathbf{6}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$
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