The length of a cuboid is 60 cm. Its height is 40% of its length and 3\4 of its width. calculate the volume of the parallelepiped.

katdoringlo
2022-09-11
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asked 2022-08-13

Finding the volume of this intersection

How to find the volume of this:

The region common to the interiors of the cylinders ${x}^{2}+{y}^{2}=1$ and ${x}^{2}+{z}^{2}=1$ and the first octant.

I tried finding the volume via double integration and the integration gets too complicated. I want to know how to solve this via triple integration. Usually there are x, y, z′s in both equations and that I can solve with triple but I dont know what to do here. I started of doing ${x}^{2}+{z}^{2}={x}^{2}+{y}^{2}$ so $z=y$ but dont know what to do with this it doesnt look like the line of intersection is $z=y$ in the diagram given.

How to find the volume of this:

The region common to the interiors of the cylinders ${x}^{2}+{y}^{2}=1$ and ${x}^{2}+{z}^{2}=1$ and the first octant.

I tried finding the volume via double integration and the integration gets too complicated. I want to know how to solve this via triple integration. Usually there are x, y, z′s in both equations and that I can solve with triple but I dont know what to do here. I started of doing ${x}^{2}+{z}^{2}={x}^{2}+{y}^{2}$ so $z=y$ but dont know what to do with this it doesnt look like the line of intersection is $z=y$ in the diagram given.

asked 2022-08-26

Finding volume of the solid

The region bounded by the curve $y=\sqrt{x}$, the x-axis and the line $x=4$ is revolved about y-axis to generate a solid. Find the volume of solid.

I found $32\pi /5$? But answer is wrong. Where did I wrong?

${\text{Volume}}_{\text{}y}=\pi {\int}_{0}^{2}[A(y){]}^{2}dy=\pi {\int}_{0}^{2}{y}^{4}dy$

The region bounded by the curve $y=\sqrt{x}$, the x-axis and the line $x=4$ is revolved about y-axis to generate a solid. Find the volume of solid.

I found $32\pi /5$? But answer is wrong. Where did I wrong?

${\text{Volume}}_{\text{}y}=\pi {\int}_{0}^{2}[A(y){]}^{2}dy=\pi {\int}_{0}^{2}{y}^{4}dy$

asked 2022-09-29

How to express volume of a sphere as a sum of infinitesimally thick discs?

I want to express the volume of a sphere with a radius r as an integral that adds up each infinitesimally thick disc within the volume. So I have $dV=A(x)dx$, where A(x) is the area of the disc that is at coordinate x. I'm having trouble finding this function in terms of r. Remember, r is the radius of the sphere. At $x=0,A=\pi {r}^{2}$, and at $x=r,A=0$. The bounds of the integral should be -r to r I believe. But what is A(x)?

I want to express the volume of a sphere with a radius r as an integral that adds up each infinitesimally thick disc within the volume. So I have $dV=A(x)dx$, where A(x) is the area of the disc that is at coordinate x. I'm having trouble finding this function in terms of r. Remember, r is the radius of the sphere. At $x=0,A=\pi {r}^{2}$, and at $x=r,A=0$. The bounds of the integral should be -r to r I believe. But what is A(x)?

asked 2022-08-19

Double Integrals and finding the volume of a pool

I am trying to find the volume of a pool and Im a little rusty on the calculus. How would I go about finding the volume of pool that is Length 10ft by Width 8ft? The depth goes from 3 to 6 feet with a slope of 3ft/10ft.

I am trying to find the volume of a pool and Im a little rusty on the calculus. How would I go about finding the volume of pool that is Length 10ft by Width 8ft? The depth goes from 3 to 6 feet with a slope of 3ft/10ft.

asked 2022-08-07

Finding volume using triple integral when i have given set.

Find the volume of:

$V=[(x,y,z):0\u2a7dz\u2a7d4-\sqrt{{x}^{2}+{y}^{2}},2x\u2a7d{x}^{2}+{y}^{2}\u2a7d4x]$

I should somehow construct triple integral here in order to solve this, which means that i have to find limits of integration for three variables, but i am just not quite sure how, i assume that i should first integrate for z since we could say that limits for z are already given, but what i am supposed to do with other two variables. When i find limits, what function i am going to integrate, is it going to be just $\iiint dzdydx$?

Find the volume of:

$V=[(x,y,z):0\u2a7dz\u2a7d4-\sqrt{{x}^{2}+{y}^{2}},2x\u2a7d{x}^{2}+{y}^{2}\u2a7d4x]$

I should somehow construct triple integral here in order to solve this, which means that i have to find limits of integration for three variables, but i am just not quite sure how, i assume that i should first integrate for z since we could say that limits for z are already given, but what i am supposed to do with other two variables. When i find limits, what function i am going to integrate, is it going to be just $\iiint dzdydx$?

asked 2022-07-21

Finding the Volume of a solid - Application of Integrals - Exercise that is not clear to understand

I've been working on a few exercises and one of them seems not clear, I'm not sure what the author meant in it. Here's the exercise:

Find the volume of the solid whose base is the area between the curve $\begin{array}{rl}y& ={x}^{3}\end{array}$ and the y axis, from $x=0$ to $y=1$, considering that his cross sections, taken perpendicular to the y axis, are squares.

I've been working on a few exercises and one of them seems not clear, I'm not sure what the author meant in it. Here's the exercise:

Find the volume of the solid whose base is the area between the curve $\begin{array}{rl}y& ={x}^{3}\end{array}$ and the y axis, from $x=0$ to $y=1$, considering that his cross sections, taken perpendicular to the y axis, are squares.

asked 2022-08-12

Volume of Solid Enclosed by an Equation

I'm having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:

Compute the volume of the solid enclosed by

1. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,x=0,y=o,z=0$

2. ${x}^{2}+{y}^{2}-2ax=0,z=0,{x}^{2}+{y}^{2}={z}^{2}$

I'm having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:

Compute the volume of the solid enclosed by

1. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,x=0,y=o,z=0$

2. ${x}^{2}+{y}^{2}-2ax=0,z=0,{x}^{2}+{y}^{2}={z}^{2}$