Question

Find the volume in the first octant bounded by the cone${\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}{\mathit{y}}^{\mathbf{2}}$ and the plane $\mathit{x}\mathbf{=}\mathbf{4}$ .

Short answer

The volume in the first octant is $\frac{16}{3}\pi$.

Step-by-step solution

Definition of double integral and mass formula

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

Draw the volume in the first octant bounded by the curve

The volume in the first octant bounded by the cone $z^2=x^2-y^2$and the plane $x=4$ .

Calculation of the volume under the curve.

Calculation integral of the volume.

$$\begin{gathered} V={\int }_0^{4}dx{\int }_0^{x}dy{\int }_0^{\sqrt{x^2-y^2}}dz \\ ={\int }_0^{4}dx{\int }_0^{x}dy\left(\sqrt{x^2-y^2}\right) \end{gathered}$$

Let $y=x\sin \theta$ then$duy=x\cos d\theta$ .

The integration bound changed to 0 and $\frac{\mathrm{\pi }}{2}$ because when $x\sin \theta =0$ the value $\theta$ will be 0 and when $y=x\sin \theta$the value $\theta$ will be $\frac{\mathrm{\pi }}{2}$.

Further calculation of the volume under the curve

Substitute changed limits to calculate limits.

$$\begin{gathered} V={\int }_0^{4}dx{\int }_0^{\pi /2}x\cos \theta d\theta \sqrt{x^2-x^2{\sin }^2\theta } \\ ={\int }_0^4{x}^{2}dx{\int }_0^{\pi /2}{\cos }^2\theta d\theta \\ ={\left.\frac{1}{4}{x}^3\right|}_0^4{\int }_0^{\pi /2}\frac{1+\cos 2\theta }{2}d\theta \end{gathered}$$

Calculation further to obtain the final value.

$$\begin{gathered} V={\left.\frac{1}{4}{x}^3\right|}_0^4{\int }_0^{\pi /2}\frac{1+\cos 2\theta }{2}d\theta \\ =\frac{56}{3}\frac{1}{2}\left[\frac{\pi }{2}+{\left.\frac{1}{2}\sin 2\theta \right|}_0^{\pi /2}\right] \\ =\frac{16}{3}\pi \end{gathered}$$

Therefore, the value is $\frac{16}{3}\pi$.