Question

Area of the circle $(x-2{)}^2+(y-3{)}^2=32$ which lies below the line y=x+1 is -
${\int }_{-2}^6\left[\right(x+1)+\sqrt{32-(x-2{)}^2}+3]dx$

Answer 1

We have,

Area of circle $={(x-2)}^2+{(y-3)}^2=32$

Line$=y=x+1$

${\int }_{-2}^6[(x+1)+\sqrt{32-{(x-2)}^2}+3]dx$

$\Rightarrow {\int }_{-2}^6[(x+1)+\sqrt{32-{(x-2)}^2}]dx+{\int }_{-2}^{6}3dx$

$\Rightarrow {\int }_{-2}^{6}xdx+{\int }_{-2}^{6}1dx+{\int }_{-2}^6\sqrt{32-{(x-2)}^2}dx+{\int }_{-2}^{6}3dx$

$\Rightarrow {\int }_{-2}^{6}xdx+{\int }_{-2}^{6}1dx+{\int }_{-2}^6{(32-{(x-2)}^2)}^{\frac{1}{2}}dx+3{\int }_{-2}^{6}1dx$

$\Rightarrow {{\left[\frac{x^2}{2}\right]}_{-2}}^6+{{\left[x\right]}_{-2}}^6+\left[\frac{{(32-{(x-2)}^2)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+3{{\left[x\right]}_{-2}}^6$

$\Rightarrow \left[\frac{6^2-{(-2)}^2}{2}\right]+[6-(-2)]+\left[\frac{{[32-{(6-2)}^2]}^{\frac{3}{2}}}{\frac{3}{2}}\right]-{\left[\frac{32-{(-2-2)}^2}{\frac{3}{2}}\right]}^{\frac{3}{2}}+3[6-(-2)]$

$\Rightarrow \left[\frac{36-4}{2}\right]+\left[8\right]+\left[\frac{{(32-16)}^{\frac{3}{2}}}{\frac{3}{2}}\right]-\left[\frac{{(32-16)}^{\frac{3}{2}}}{\frac{3}{2}}\right]+3\times 8$

$\Rightarrow \frac{32}{2}+8+24$

$\Rightarrow 16+8+24$

$\Rightarrow 48\text{units}$

Hence, this is the answer.