# How do you find the volume of the solid bounded by x^2+y^2=4 and z=x+y in the first octant?

Jul 16, 2016

16/3

#### Explanation:

you'll want to use

$\int \int z \left(x , y\right) \mathrm{dA}$

$= {\int}_{y = 0}^{2} {\int}_{x = 0}^{\sqrt{4 - {y}^{2}}} \left(x + y\right) \setminus \mathrm{dx} \setminus \mathrm{dy}$

$= {\int}_{y = 0}^{2} {\left[{x}^{2} / 2 + x y\right]}_{x = 0}^{\sqrt{4 - {y}^{2}}} \setminus \mathrm{dy}$

$= {\int}_{y = 0}^{2} 2 - {y}^{2} / 2 + y \sqrt{4 - {y}^{2}} \setminus \mathrm{dy}$

$= {\left[2 y - {y}^{3} / 6 - \frac{1}{3} {\left(4 - {y}^{2}\right)}^{\frac{3}{2}}\right]}_{y = 0}^{2}$

$= \frac{16}{3}$