written 5.0 years ago by |
Solution:
Paraboloid: $x^{2}+y^{2}=4 z \quad$ Plane $: z=4$
$(x, y, z) \quad \longrightarrow \quad(r, \theta, z)$
Put $x=r \cos \theta, y=r \sin \theta, z=z \quad \therefore x^{2}+y^{2}=r^{2}$
$\therefore$ Paraboloid: $r^{2}=4 z \quad$ and $\quad$ Plane $: z=4$
If we are passing one arrow parallel to z axis from -ve to to we will get limits of z
$\begin{aligned} \therefore \quad \frac{r^{2}}{4} & \leq z \leq 4 \\ 0 & \leq r \leq 4 \\ & 0 \leq \theta \leq \frac{\pi}{2} \end{aligned}$
Volume of given paraboloid cut off by the plane is given by
$\begin{aligned} V &=4 \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} \int_{\frac{r^{2}}{4}}^{4} r \ d r \ d \theta \ d z \\ &=4 \int_{0}^{\frac{\pi}{2}} \int_{0}^{4}\left[4 r-\frac{r^{4}}{16}\right]^4_{\frac{r^2}{4}} d r \ d \theta \end{aligned}$
$=4 \int_{0}^{\frac{\pi}{2}} \int_{0}^{4}\left(4 r-\frac{r^{3}}{4}\right) d r \ d \theta$
$=4 \int_{0}^{\frac{\pi}{2}}\left[2 r^{2}-\frac{r^{4}}{16}\right]_{0}^{4} \mathrm{d} \theta$
$=4 \int_{0}^{\frac{\pi}{2}}(32-16) \mathrm{d} \theta$
$V=32 \pi$ cubic units