What is the volume of a right right pyramid whose base is a square with a side 6m long and whose altitude is aqual to base side? wich one is A 36M OR B 72M OR C 108M OR D 216M 3 SQUARE WICH ONE IS IT

Darryl English
2022-07-26
Answered

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Damarion Pierce

Answered 2022-07-27
Author has **11** answers

The volume of a right pyramid is:

b*h/3

where b is the area of the base.

B = 6*6 = 36

h = 6

$36\ast 6/3=72{m}^{3}$

b*h/3

where b is the area of the base.

B = 6*6 = 36

h = 6

$36\ast 6/3=72{m}^{3}$

asked 2022-07-15

Finding volume of the region enclosed by ${x}^{2}+{y}^{2}-6x=0,z=\sqrt{36-{x}^{2}-{y}^{2}}$

$z=0$

I'm trying to calculate the volume of the region enclosed by the cylinder ${x}^{2}+{y}^{2}-6x=0$, the semicircle $z=\sqrt{36-{x}^{2}-{y}^{2}}$ and the plane $z=0$. I tried moving the region towards -x so that the cylinder has it's center at zero. I ended up using these equations and proposing this integral using cylindrical coordinates:

New cylinder ${x}^{2}+{y}^{2}=9$

New semicircle $z=\sqrt{36-(x+3{)}^{2}-{y}^{2}}$

${\int}_{0}^{2\pi}{\int}_{0}^{3}{\int}_{0}^{\sqrt{36-(rcos(\theta )+3{)}^{2}-(rsin(\theta ){)}^{2}}}r\cdot dzdrd\theta $

For some reason I can't solve it this way and I don't know what am I doing wrong.

$z=0$

I'm trying to calculate the volume of the region enclosed by the cylinder ${x}^{2}+{y}^{2}-6x=0$, the semicircle $z=\sqrt{36-{x}^{2}-{y}^{2}}$ and the plane $z=0$. I tried moving the region towards -x so that the cylinder has it's center at zero. I ended up using these equations and proposing this integral using cylindrical coordinates:

New cylinder ${x}^{2}+{y}^{2}=9$

New semicircle $z=\sqrt{36-(x+3{)}^{2}-{y}^{2}}$

${\int}_{0}^{2\pi}{\int}_{0}^{3}{\int}_{0}^{\sqrt{36-(rcos(\theta )+3{)}^{2}-(rsin(\theta ){)}^{2}}}r\cdot dzdrd\theta $

For some reason I can't solve it this way and I don't know what am I doing wrong.

asked 2022-08-19

Finding volume of the solid of revolution

The function is:

$y=\mathrm{sin}({x}^{2})$ and the boundaries are $y=0$, $x=0$, and $x=\sqrt{\pi}$.

The function is rotated around the y-axis, and since the functions are difficult to express in terms of y and integrate with respect to dy, I use the formula of $V=2\pi {\int}_{0}^{\sqrt{\pi}}\mathrm{sin}({x}^{2})dx$

My two questions here are:

1) What method can I use to solve the integral of $\mathrm{sin}({x}^{2})$?

2) How do I limit the boundary in some way to not include the area below the x-axis?

The function is:

$y=\mathrm{sin}({x}^{2})$ and the boundaries are $y=0$, $x=0$, and $x=\sqrt{\pi}$.

The function is rotated around the y-axis, and since the functions are difficult to express in terms of y and integrate with respect to dy, I use the formula of $V=2\pi {\int}_{0}^{\sqrt{\pi}}\mathrm{sin}({x}^{2})dx$

My two questions here are:

1) What method can I use to solve the integral of $\mathrm{sin}({x}^{2})$?

2) How do I limit the boundary in some way to not include the area below the x-axis?

asked 2022-07-16

Finding the volume by Shell method: $y={x}^{2},y=2-{x}^{2}$, $y={x}^{2},y=2-{x}^{2}$ about the line $x=1$.

what I get from this after graphing is:

$2\pi \int (1-x)(2-2{x}^{2})dx$

which becomes: $2\pi \int (2-2{x}^{2}-2x+2{x}^{3})dx$

integrating that I get:

$2\pi [2x-\frac{2}{3}{x}^{3}-{x}^{2}+\frac{1}{2}{x}^{4}]$ from 0 to 1

My answer is $\frac{5}{3}\pi $ but my book says the answer is: $\frac{16}{3}\pi $

Could someone tell me where I went wrong? Was it the upper lower bounds?

what I get from this after graphing is:

$2\pi \int (1-x)(2-2{x}^{2})dx$

which becomes: $2\pi \int (2-2{x}^{2}-2x+2{x}^{3})dx$

integrating that I get:

$2\pi [2x-\frac{2}{3}{x}^{3}-{x}^{2}+\frac{1}{2}{x}^{4}]$ from 0 to 1

My answer is $\frac{5}{3}\pi $ but my book says the answer is: $\frac{16}{3}\pi $

Could someone tell me where I went wrong? Was it the upper lower bounds?

asked 2022-08-31

About finding the rotational volume

A closed area of y-axis, $y=\pi ,y=x+sin(x)$ is rotated on line $y=x$. Find the volume of it.

So, I think I have to do an integral looks like this.

$V={\int}_{0}^{\sqrt{2}\pi}\pi {r}^{2}dt$, where r is the radius of disk and t is a variable on line $y=x$.

How should I set up the integral and change the variables?

A closed area of y-axis, $y=\pi ,y=x+sin(x)$ is rotated on line $y=x$. Find the volume of it.

So, I think I have to do an integral looks like this.

$V={\int}_{0}^{\sqrt{2}\pi}\pi {r}^{2}dt$, where r is the radius of disk and t is a variable on line $y=x$.

How should I set up the integral and change the variables?

asked 2022-11-12

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Using Symmetry for finding volume

I have a confusion regarding the symmetry of the volume in the following question.

Find the volume common to the sphere ${x}^{2}+{y}^{2}+{z}^{2}=16$ and cylinder ${x}^{2}+{y}^{2}=4y$.

The author used polar coordinates $x=rcos\theta $ and $y=rsin\theta $ and does something like this:

Required volume $V=4{\int}_{0}^{\pi /2}{\int}_{0}^{4sin\theta}(16-{r}^{2}{)}^{1/2}rdrd\theta $. The reason for multiplying by 4 is the symmetry of the solid w.r.t. xy-plane.

My point of confusion is that this solid cannot be cut into 4 identical parts, so how it can be multiplied by 4?

I have a confusion regarding the symmetry of the volume in the following question.

Find the volume common to the sphere ${x}^{2}+{y}^{2}+{z}^{2}=16$ and cylinder ${x}^{2}+{y}^{2}=4y$.

The author used polar coordinates $x=rcos\theta $ and $y=rsin\theta $ and does something like this:

Required volume $V=4{\int}_{0}^{\pi /2}{\int}_{0}^{4sin\theta}(16-{r}^{2}{)}^{1/2}rdrd\theta $. The reason for multiplying by 4 is the symmetry of the solid w.r.t. xy-plane.

My point of confusion is that this solid cannot be cut into 4 identical parts, so how it can be multiplied by 4?

asked 2022-07-14

Gauss-divergence theorem for volume integral of a gradient field

I need to make sure that the derivation in the book I am using is mathematically correct. The problem is about finding the volume integral of the gradient field. The author directly uses the Gauss-divergence theorem to relate the volume integral of gradient of a scalar to the surface integral of the flux through the surface surrounding this volume, i.e.

${\int}_{CV}^{}\mathrm{\nabla}\varphi dV={\int}_{\delta CV}^{}\varphi d\mathbf{S}$

I need to make sure that the derivation in the book I am using is mathematically correct. The problem is about finding the volume integral of the gradient field. The author directly uses the Gauss-divergence theorem to relate the volume integral of gradient of a scalar to the surface integral of the flux through the surface surrounding this volume, i.e.

${\int}_{CV}^{}\mathrm{\nabla}\varphi dV={\int}_{\delta CV}^{}\varphi d\mathbf{S}$

asked 2022-09-20

What is the volume in cubic inches of a box that is 25 cm by 25 cm by 25 cm (given 1 inch=2.5 cm approx.)