Find the centroid of the solid in part (a).

In summary, the conversation is about finding the volume and centroid of a solid above a cone and below a sphere, with given equations. The volume is found to be 10pi and the centroid is calculated as (0,0,21/40). However, the book's answer for the inner integration of rho differs, leading to confusion and discussion about the correct answer. Ultimately, it is discovered that a simple mistake was made in the calculation.
  • #1
Litcyb
36
0

Homework Statement



Find the volume of the solid that lies above the cone
ϕ=pi/3
and below the sphere
ρ=4cosϕ
.

Homework Equations



Find the centroid of the solid in part (a)..

The Attempt at a Solution



For the volume I got 10pi which I am fairly sure is correct. I attempted trying to calculate the centroid and i got a different answer from the book. I have (0,0,21/40)

and I would like to know what did i do incorrect. I calculated ybar and xbar and i got 0.
for zbar, i did the following

∫0→pi/3∫0→2pi∫0→4cosϕ (ρ^3cosϕsinϕdρdθdϕ)the book's answer for the inner integration of rho was ∫0→2pi∫0→pi/3 (cosϕsinϕ(64cos^4(ϕ))dϕdθ

is this correct? isn't supposed to be ∫0→2pi∫0→pi/3 (cosϕsinϕ(64cos^4(ϕ)/4)dϕdθ?

or am I wrong? (I divided by 4, because that's the rule of integration)
 
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  • #2


Litcyb said:

Homework Statement



Find the volume of the solid that lies above the cone
ϕ=pi/3
and below the sphere
ρ=4cosϕ
.

Homework Equations



Find the centroid of the solid in part (a).


.

The Attempt at a Solution



For the volume I got 10pi which I am fairly sure is correct. I attempted trying to calculate the centroid and i got a different answer from the book. I have (0,0,21/40)

and I would like to know what did i do incorrect. I calculated ybar and xbar and i got 0.
for zbar, i did the following

∫0→pi/3∫0→2pi∫0→4cosϕ (ρ^3cosϕsinϕdρdθdϕ)


the book's answer for the inner integration of rho was ∫0→2pi∫0→pi/3 (cosϕsinϕ(64cos^4(ϕ))dϕdθ

is this correct? isn't supposed to be ∫0→2pi∫0→pi/3 (cosϕsinϕ(64cos^4(ϕ)/4)dϕdθ?

or am I wrong? (I divided by 4, because that's the rule of integration)

You get (4cosϕ)^4/4, right? What's 4^4/4?
 
  • #3


hahaha!I fell so dumb now. Making very stupid mistakes. thanks for output!
 

FAQ: Find the centroid of the solid in part (a).

What is the centroid of a solid?

The centroid of a solid is the point where all the mass of the solid can be considered to be concentrated. It is the geometric center of the solid and is a useful concept in physics and engineering.

How do you find the centroid of a solid?

To find the centroid of a solid, you need to calculate the average of the coordinates of all the points that make up the solid. This can be done using either geometric formulas or integration techniques, depending on the complexity of the solid.

What is the importance of finding the centroid of a solid?

Knowing the location of the centroid is important in determining the stability, balance, and strength of a solid. It is also used in calculating the moments of inertia and performing other engineering calculations.

Can the centroid of a solid be located outside of the solid?

Yes, the centroid of a solid can be located outside of the solid, especially in cases where the solid has an irregular shape or varying density. In these cases, the centroid may be located at a point where there is no actual mass present.

How is the centroid of a solid different from the center of mass?

The centroid of a solid is the point where all the mass of the solid is concentrated, while the center of mass is the point where the weight of the solid is balanced. In most cases, the centroid and center of mass will be at the same location, but in some cases, such as a hollow object, they may be different.

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