Marenonigt

2021-12-18

Consider the solid in xyz-space, which contains all points (x, y, z) whose z-coordinate satisfies
$0\le z\le 4-{x}^{2}-{y}^{2}$
Which statements do hold?
a) The solid is a sphere
b) The solid is apyramid
c) Its volume is $8\pi$
d) Its volume is $\frac{16\pi }{3}$

Carl Swisher

Expert

Step 1
$\int \int \int dv$
$⇒\int \int \int dzdydx$
$z=0$ to $z=4-{x}^{2}-{y}^{2}$
$⇒\int \int \left({\int }_{0}^{4-{x}^{2}-{y}^{2}}dz\right)dydx$
$⇒\int \int \left(4-{x}^{2}-{y}^{2}\right)dydx$
Change in folar co-ordinates, tre get
$x=r\mathrm{cos}\theta$
$y=r\mathrm{sin}\theta$
$r=0$ to $r=2$
$dxdy=rdrd\theta$
$\theta =0$ to $\theta =2\pi$
$⇒{\int }_{0}^{2\pi }{\int }_{0}^{2}\left(4-{r}^{2}\right)rdrd\theta$
Let $4-{r}^{2}=t⇒rdr=\frac{-dt}{2}$
$t=4$ to $t=4-4=0$
$⇒{\int }_{0}^{2\pi }\left({\int }_{4}^{0}-t\frac{dt}{2}\right)d\theta ⇒\frac{1}{2}{\int }_{0}^{2\pi }\left({\int }_{0}^{4}tdt\right)d\theta$
$⇒\frac{1}{2}{\int }_{0}^{2\pi }{\left(\frac{{t}^{2}}{2}\right)}_{0}^{4}d\theta ⇒\frac{1}{2}{\int }_{0}^{2\pi }8d\theta$
$⇒4{\int }_{0}^{2\pi }d\theta ⇒4\left(2\pi \right)$
$⇒2\pi$

Carl Swisher

Expert

Here, take limit: $0\le 4-{x}^{2}-{y}^{2}$
${x}^{2}+{y}^{2}\le 4$
And we know,
${\left(4-{x}^{2}-{y}^{2}\right)}_{max}=4$ when
So, $0\le z\le 4$
For each value of $z\ge 0$, it will be a disk of at each value of $z\in R$.
Now, volume in $=\frac{\pi }{B}{r}^{2}h$
$=\frac{\pi }{B}×{\left(2\right)}^{2}×\left(4\right)=16\frac{\pi }{B}$

nick1337

Expert

Step 1
Consider the solid  in xyz-space which contains all points (x,y,z) whose z-coordinate satisfies $0\le z\le 4-{x}^{2}-{y}^{2}$
Step 2
The solid is paraboloid.
We use cylindrical coordinates

$x=r\mathrm{cos}t,y=r\mathrm{sin}t,0\le z\le 4-{r}^{2},0\le t\le 2\pi$
and
$⇒{r}^{2}=4$
$⇒r=2$
$⇒0\le r\le 2$
The volume of the solid is given by
$\begin{array}{}V=\int \int {\int }_{S}dV\\ ={\int }_{0}^{2}{\int }_{0}^{2\pi }{\int }_{0}^{4-{r}^{2}}rdzdtdr\\ ={\int }_{0}^{2}{\int }_{0}^{2\pi }|z{|}_{0}^{4-{r}^{2}}rdtdr\\ ={\int }_{0}^{2}{\int }_{0}^{2\pi }\left(4-{r}^{2}\right)rdtdr\\ ={\int }_{0}^{2}r\left(4-{r}^{2}\right)\left[t{\right]}_{0}^{2\pi }dr\\ ={\int }_{0}^{2}r\left(4-{r}^{2}\right)\left[2\pi -0\right]dr\\ =2\pi {\int }_{0}^{2}\left(4r-{r}^{3}\right)dr\\ =2\pi \left[2{r}^{2}-\frac{{r}^{4}}{4}{\right]}_{0}^{2}\\ =2\pi \left[2×{2}^{2}-\frac{{2}^{4}}{4}-0\right]\\ =2\pi \left[8-4\right]\\ =8\pi \end{array}$
Thus, volume of the solid is $8\pi$.
Hence, option (c) is correct.

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