Find V of a solid made by rotating line on an axis

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Homework Help Overview

The discussion revolves around finding the volume of a solid generated by rotating the line defined by the equation x=4tan(y*pi/3) about the y-axis, specifically within the bounds of 0≤y≤1.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the volume integral and the interpretation of the radius as a function of y. There are attempts to simplify the integral using trigonometric identities and substitutions. Questions arise regarding the correctness of integration steps and the final evaluation of the integral.

Discussion Status

Some participants have offered guidance on simplifying the integral and have pointed out potential errors in the integration process. There is an ongoing exploration of different methods to approach the problem, including the use of trigonometric identities.

Contextual Notes

One participant notes the challenge of understanding the final steps of the integration process, indicating a reliance on external resources for clarification. There is also mention of a lack of assistance from a teacher, highlighting the collaborative nature of the discussion.

TomFoolery
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Homework Statement



If a solid is generated by rotating the line (x=4tan(y*pi/3)) on the y-axis. Find the volume between the area 0≤y≤1.

Homework Equations



I know that, when slicing a section (A(x)), I will generate a circle. This gives me two of the dimensions (by using area of a circle formula; A=pi*r2).

r=4tan(y*pi/3)

Therefore, V= [itex]\int[/itex] pi*(4tan(y*pi/3))2
Lower Lim: 0, Upper lim:1

The Attempt at a Solution



This should become:

V = 16*pi [itex]\int[/itex] (tan(y*pi/3)2

Then,
u = tan (y*pi/3)
w = y*pi/3
V = 16*pi [itex]\int[/itex] (u2)(tan(w))(y*pi/3)

then,
V = (16pi) (tan(pi*y/3)3/3) (ln|cos(pi*y/3)|) (pi*y2/6)


My calculator tells me i should end up with 4(3^(1/2) - pi) but I have no idea how. It's just this last step that is killing me.
 
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Your integration is not correct. You have the right idea (the radius is the function while the height is dy), but the integral is more elementary than you are making it. Your integral is,
[itex]16\pi\int_{0}^{1}tan^2\frac{y\pi}{3}dy[/itex]
Consider a trig identity to make this integral easy to do rather than using a substitution. Let us know if you need anymore help.
 
Got it, it's because Tan2 = sec2(x)-1

this allows you to integrate to get:

(16pi) [(tan(pi*y/3)) (3/pi) -y]

Since the lower limit is 0, the tan part becomes 0 so the answer is F(b) - 0 = F(b)

Thank you for your help, my teacher isn't much assistance so I'm learning how to use the rest of the world as a teacher this semester.
 
TomFoolery said:

Homework Statement



If a solid is generated by rotating the line (x=4tan(y*pi/3)) on the y-axis. Find the volume between the area 0≤y≤1.

Homework Equations



I know that, when slicing a section (A(x)), I will generate a circle. This gives me two of the dimensions (by using area of a circle formula; A=pi*r2).

r=4tan(y*pi/3)

Therefore, V= [itex]\int[/itex] pi*(4tan(y*pi/3))2
Lower Lim: 0, Upper lim:1

The Attempt at a Solution



This should become:

V = 16*pi [itex]\int[/itex] (tan(y*pi/3)2

Then,
u = tan (y*pi/3)
w = y*pi/3
V = 16*pi [itex]\int[/itex] (u2)(tan(w))(y*pi/3)

then,
V = (16pi) (tan(pi*y/3)3/3) (ln|cos(pi*y/3)|) (pi*y2/6)


My calculator tells me i should end up with 4(3^(1/2) - pi) but I have no idea how. It's just this last step that is killing me.

Google "Pappus's Theorem".

RGV
 

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