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

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SUMMARY

The volume of a solid generated by rotating the line defined by the equation x=4tan(yπ/3) around the y-axis from y=0 to y=1 is calculated using the integral V = 16π∫(tan²(yπ/3))dy. The correct evaluation of this integral simplifies to V = 16π[(tan(πy/3))(3/π) - y] evaluated from 0 to 1. The final result is V = 4(√3 - π). The discussion emphasizes the importance of using trigonometric identities to simplify the integration process.

PREREQUISITES
  • Understanding of integral calculus, specifically volume of revolution.
  • Familiarity with trigonometric functions and identities, particularly tan and sec.
  • Knowledge of the area of a circle formula, A=πr².
  • Ability to perform definite integrals and evaluate limits.
NEXT STEPS
  • Study trigonometric identities to simplify integrals, focusing on tan² and sec².
  • Learn about the method of volumes of revolution in calculus.
  • Practice evaluating definite integrals involving trigonometric functions.
  • Explore Pappus's Theorem for further insights into volumes of solids of revolution.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques and applications in finding volumes of solids of revolution.

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

The Attempt at a Solution



This should become:

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

Then,
u = tan (y*pi/3)
w = y*pi/3
V = 16*pi \int (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,
16\pi\int_{0}^{1}tan^2\frac{y\pi}{3}dy
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= \int pi*(4tan(y*pi/3))2
Lower Lim: 0, Upper lim:1

The Attempt at a Solution



This should become:

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

Then,
u = tan (y*pi/3)
w = y*pi/3
V = 16*pi \int (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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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