Integration of Logs HELP (1 Viewer)

[sando]

1.. Find the area between the curve y= In x, the y-axis and the lines y = 2 and y = 4


2.. Find the exact volume of the solid formed when the curve y = log_ex is rotated about the y-axis from y = 1 to y = 3

 

[SoulSearcher]

1. y = ln x
x = e^y
.'. int. [e^y](2->4) dy
= [e^y](2->4)
= (e⁴ - e²) units²

2. y = ln x
x = e^y
x² = e^2y
.'. pi * int. [e^2y](1->3)
= pi * [e^2y/2](1->3)
= pi/2 * [e^2y](1->3)
= pi/2 * (e⁶-e²) units³

 

[who_loves_maths]

Hey sando,

Question 1:

The equation of interest is ln(x) = y ---> x = e^y ,
because you will be integrating with respect to dy:

I = {4 -> 2}∫e^y dy
= [e^y]{4 -> 2} = e⁴ - e² = e²(e² - 1) units²


Question 2:

Once again, the equation of interest is x = e^y for the same above reason:

Using the standard integral form for solids of revolutions: V = pi∫x² dy

V = pi.{3 -> 1}∫e^2y dy
= (pi/2).[e^2y]{3 -> 1}
= (pi/2).(e⁶ - e²)
= ((pi.e²)/2)(e⁴ - 1) units³


Hope that helps.


EDIT: Acknowledgement re SoulSearcher.

 

[suebear]

1.. Find the area between the curve y= In x, the y-axis and the lines y = 2 and y = 4

1st u gotta make x the subject
x = e to power of y

A = integral b/w 4 and 2 of ey
= [ln y ] with 4 at the top and 2 at the bottom
= ln 4 - ln 2 units sq

2.. Find the exact volume of the solid formed when the curve y = log<SUB>e</SUB>x is rotated about the y-axis from y = 1 to y = 3

again make x the subject
x= e to power of y

v= pie * integral of x(sqd) (*=times)
= pie * integral of e2y with 3 at top and 2 at bottom
=pie *[1/2 e 2y] b/w 3 and 2 (thats e to power 2y)
= pie * 1/2 ( e 6 - e 4)

i hope that was helpful...its sooo hard doin maths on a comp
good luk



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[sando]

for Q 2.. why pi/2 ???

 

[SoulSearcher]

When you integrate e^2y, you get e^2y/2 + C. The half has simply been taken out of the integral to make the calculations easier.

 

[pLuvia]

When you integrate x=e^2y, it equals to 1/2e^2y

 

[sando]

ahh yes.. thanks.. its so hard to understand on computer

 

[sando]

another question:

Find the area enclosed by the curve y = 1-e^-x, the x-axis, and the lines x = -2 and x = 2

 

[pLuvia]

int.{-2 to 2}(1-e^-x)dx
=[x+e^-x]{-2 to 2}
=(2+e^-2)-(-2+e²)
=4+e^-2-e²

 

[sando]

thanks