# Consider the solid in xyz-space, which contains all points (x,

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

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Carl Swisher

Step 1
$$\displaystyle\int\int\int{d}{v}$$
$$\displaystyle\Rightarrow\int\int\int{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle{z}={0}$$ to $$\displaystyle{z}={4}-{x}^{{{2}}}-{y}^{{{2}}}$$
$$\displaystyle\Rightarrow\int\int{\left({\int_{{{0}}}^{{{4}-{x}^{{{2}}}-{y}^{{{2}}}}}}{\left.{d}{z}\right.}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow\int\int{\left({4}-{x}^{{{2}}}-{y}^{{{2}}}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
Change in folar co-ordinates, tre get
$$\displaystyle{x}={r}{\cos{\theta}}$$
$$\displaystyle{y}={r}{\sin{\theta}}$$
$$\displaystyle{r}={0}$$ to $$\displaystyle{r}={2}$$
$$\displaystyle{\left.{d}{x}\right.}{\left.{d}{y}\right.}={r}{d}{r}{d}\theta$$
$$\displaystyle\theta={0}$$ to $$\displaystyle\theta={2}\pi$$
$$\displaystyle\Rightarrow{\int_{{{0}}}^{{{2}\pi}}}{\int_{{{0}}}^{{{2}}}}{\left({4}-{r}^{{{2}}}\right)}{r}{d}{r}{d}\theta$$
Let $$\displaystyle{4}-{r}^{{{2}}}={t}\Rightarrow{r}{d}{r}={\frac{{-{\left.{d}{t}\right.}}}{{{2}}}}$$
$$\displaystyle{t}={4}$$ to $$\displaystyle{t}={4}-{4}={0}$$
$$\displaystyle\Rightarrow{\int_{{{0}}}^{{{2}\pi}}}{\left({\int_{{{4}}}^{{{0}}}}-{t}{\frac{{{\left.{d}{t}\right.}}}{{{2}}}}\right)}{d}\theta\Rightarrow{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}\pi}}}{\left({\int_{{{0}}}^{{{4}}}}{t}{\left.{d}{t}\right.}\right)}{d}\theta$$
$$\displaystyle\Rightarrow{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}\pi}}}{{\left({\frac{{{t}^{{{2}}}}}{{{2}}}}\right)}_{{{0}}}^{{{4}}}}{d}\theta\Rightarrow{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}\pi}}}{8}{d}\theta$$
$$\displaystyle\Rightarrow{4}{\int_{{{0}}}^{{{2}\pi}}}{d}\theta\Rightarrow{4}{\left({2}\pi\right)}$$
$$\displaystyle\Rightarrow{2}\pi$$

###### Not exactly what you’re looking for?
Carl Swisher

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

nick1337

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

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