Question

Find the volume in the first octant bounded by the paraboloid $\mathit{z}\mathbf{=}\mathbf{1}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{-}{\mathit{y}}^{\mathbf{2}}$, the plane $\mathit{x}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{1}$, and all three coordinate planes.

Short answer

The volume in the first octant is$\frac{1}{3}\text{.}$

Step-by-step solution

Definition of double integral and mass formula

The double integralof $f(x,y)$over the area A in the$(x,y)$ plane as the limit of this sum, and write it as ${\iint }_{A}f(x,y)dxdy$.

Drawing the volume in the first octant bounded by the curve

The volume in the first octant is bounded by the paraboloid $z=1-x^2-y^2$, the plane $x+y=1$, and all three coordinate planes.

Calculation of the volume under the curve

Calculation integral of the volume.

$$\begin{gathered} V={\int }_0^{1}dx{\int }_0^{1-x}dy{\int }_0^{1-x^2-y^2}dz \\ ={\int }_0^{1}dx{\int }_0^{1-x}dy\left(1-x^2-y^2\right) \\ ={\left.{\int }_0^{1}dx\left(y-x^{2}y-\frac{1}{3}{y}^3\right)\right|}_0^{1-x} \\ ={\int }_0^1\left((1-x)-x^2(1-x)-\frac{1}{3}{(1-x)}^3\right)dx \end{gathered}$$

Substitute,$u=1-x$, then $du=-dx$.

Further calculation of the volume under the curve

The volume is given by,

$$\begin{gathered} V=-{\int }_1^{0}du\left(u-{(1-u)}^{2}u-\frac{1}{3}{u}^3\right) \\ ={\int }_0^1\left(-\frac{4}{3}{u}^3+2u^2\right)du \\ ={\left.\left[-\frac{1}{3}{u}^4+\frac{2}{3}{u}^3\right]\right|}_0^1 \\ =\frac{1}{3} \end{gathered}$$

Therefore, the value is $\frac{1}{3}\text{.}$,