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

Marenonigt 2021-12-18 Answered
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 \(\displaystyle{\frac{{{16}\pi}}{{{3}}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Carl Swisher
Answered 2021-12-19 Author has 4759 answers

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?
Ask My Question
0
 
Carl Swisher
Answered 2021-12-20 Author has 4759 answers

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 \(\displaystyle={\frac{{\pi}}{{{B}}}}{r}^{{{2}}}{h}\)
\(\displaystyle={\frac{{\pi}}{{{B}}}}\times{\left({2}\right)}^{{{2}}}\times{\left({4}\right)}={16}{\frac{{\pi}}{{{B}}}}\)

0
nick1337
Answered 2021-12-28 Author has 9671 answers

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.

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-12-21
A. Let \(\displaystyle{\left({X},\ {d}\right)}\) be a metric space. Define a diameter of a subset A of X and then the diameter of an open ball with center at \(\displaystyle{x}_{{{0}}}\) and radios \(\displaystyle{r}{>}{0}\)
B. Use (A) to show that every convergence sequence in \(\displaystyle{\left({X},\ {d}\right)}\) is bounded.
asked 2021-12-17
Find the distance d between the points \(\displaystyle{P}_{{{1}}}\) and \(\displaystyle{P}_{{{2}}}\)
\(\displaystyle{P}_{{{1}}}={\left({1.2},\ {2.3}\right)}\)
\(\displaystyle{P}_{{{2}}}={\left(-{0.3},\ {1.1}\right)}\)
asked 2021-12-01
Find the points on the ellipse \(\displaystyle{4}{x}^{{{2}}}+{y}^{{{2}}}={4}\) that are the farthest from the point (1;0)
asked 2021-12-11
Consider the sequence \(\displaystyle{\left\lbrace{A}_{{{n}}}\right\rbrace}\), whose nth term is given by \(\displaystyle{A}_{{{n}}}={\left({1}+{\frac{{{2}}}{{{n}}}}\right)}^{{{3}{n}}}\)
Use limits to analyticlly show how we can determine whether the sequence is convergent or divergent. If it is convergent, find the limit of the sequence. Write a sentence to summariae your findings.
asked 2021-12-13
Let \(\displaystyle{Q}^{{{2}}}\) be the rational plane of all ordered pairs (x,y) of rational numbers with the usual interpretations of the undefined geometric terms used in analytical geometry. Show that axiom C-1 and the elementary continuity principle fail in \(\displaystyle{Q}^{{{2}}}\). (Hint: the segment from (0,0) to (1,1) can not be laid off on the x axis from the origin.
Axiom C-1: If A, B are two points on a line a, and if A' is a point upon the same or another line a′ , then, upon a given side of A′ on the straight line a′ , we can always find a point B′ so that the segment AB is congruent to the segment A′B′ . We indicate this relation by writing \(\displaystyle{A}{B}\stackrel{\sim}{=}{A}′{B}′\). Every segment is congruent to itself; that is, we always have \(\displaystyle{A}{B}\stackrel{\sim}{=}{A}{B}\).
asked 2021-12-18

Consider the case of a rotating wheel at rest and starting a clockwise rotation, meaning the negative direction of the angular velocity, and increasing (negatively) its value up to \(\displaystyle-{12}{r}{a}\frac{{d}}{{\sec{}}}\) for 2 seconds. It then maintains a constant velocity for 2 seconds, and then uniformly reduces the magnitude of the velocity for 2 seconds until the wheel is momentarily stopped and restarts its rotation counter clockwise with positive angular velocity, accelerating up to \(\displaystyle{20}{r}{a}\frac{{d}}{{\sec{}}}\) in 2 seconds and remaining at a constant rotation for 2 more seconds. Finally, the wheel stops gradually in 2 seconds. Next you can see the graph of angular velocity versus time of this rotation:
\[\begin{array}{|c|c|}\hline \text{Slope:} & \\ \hline \text{Origin intercept:} & \\ \hline \text{Equation:} \omega=f(t) \end{array}\]
image

asked 2020-12-25

For the V vector space contains all \(\displaystyle{2}\times{2}\) matrices. Determine whether the \(\displaystyle{T}:{V}\rightarrow{R}^{{1}}\) is the linear transformation over the \(\displaystyle{T}{\left({A}\right)}={a}\ +\ {2}{b}\ -\ {c}\ +\ {d},\) where \(A=\begin{bmatrix}a & b \\c & d \end{bmatrix}\)

...