integration qs (1 Viewer)

[ToO LaZy ^*]

1.
sketch on the same diagram, the curves y = cosx, y = cos^2 x for 0<=x<=pi/2
find the area enclosed betweem these curves and the volume generated when this area is rotated about the x-axis.

i got this really long answer, and i think it's wrong, could someone help me with this one (don't worry about the sketch bit)


2.
the area between the curve y = sin^2 x and the x-axis between x=0 and x=pi/2 is rotated through one complete revolution about the x-axis

i) use the simpson's rule with 3 function values to find an approximation to the volume of the solid of revolution, leaving the answer in terms of pi.

 

[Calculon]

/pi/2................./pi/2
pi|cos^4 x dx - .pi|cos^2 x
/0....................../0

cos[2x]=2cos^2 x -1
(cos[2x] + 1)/2 = cos^2 x

cos^4 x = (cos^2[2x] + 2cos[2x] +1)/4
= (cos4x + 1)/8 +(2cos[2x]+1)/4

/pi/2................................................./pi/2
pi|(cos4x + 1)/8 +(2cos[2x]+1)/4 - .pi|(cos[2x] + 1)/2
/0...................................................../0

pi[sin4x/32 + x/8 + sin2x/4 + x/4 - sin2x/4 -x/2](pi/2-->0)

pi[pi/16+pi/8 - pi/4]
pi((2pi+pi-4pi))/16
-pi²/16
and find the absolute value
pi²/16

Now that answer is most likely riddled with mistakes

 

[ToO LaZy ^*]

Originally posted by Calculon
/pi/2................./pi/2

/0....................../0

what do those mean?..

 

[ToO LaZy ^*]

Originally posted by Calculon

pi[pi/16+pi/8 - pi/4]

oh...and where did the -pi/4 come from..
when you sub x = 0 into pi[sin4x/32 + x/8 + sin2x/4 + x/4 - sin2x/4 -x/2]..don't you get 0?

 

[ToO LaZy ^*]

Originally posted by ToO LaZy ^*
1.
sketch on the same diagram, the curves y = cosx, y = cos^2 x for 0<=x<=pi/2
find the area enclosed betweem these curves and the volume generated when this area is rotated about the x-axis.

i got this really long answer, and i think it's wrong, could someone help me with this one (don't worry about the sketch bit)


2.
the area between the curve y = sin^2 x and the x-axis between x=0 and x=pi/2 is rotated through one complete revolution about the x-axis

i) use the simpson's rule with 3 function values to find an approximation to the volume of the solid of revolution, leaving the answer in terms of pi.

^BUMP
help?

 

[kpq_sniper017]

Originally posted by ToO LaZy ^*
2.
the area between the curve y = sin^2 x and the x-axis between x=0 and x=pi/2 is rotated through one complete revolution about the x-axis

i) use the simpson's rule with 3 function values to find an approximation to the volume of the solid of revolution, leaving the answer in terms of pi.

y=sin²x
.'. y²=sin⁴x

V=pi ₀}^pi/2 sin⁴x dx

Using Simpson's Rule:
h=(pi/2-0)/2
=pi/4

.'. V ~ pi x (pi/4)/3 [f(0) + 4f(pi/4) + f(pi/2)]
= pi²/12 x [0 + 1 + 1]
= pi²/12 x 2
~ pi²/6 units³

~ is approx. equal to


I'm pretty sure that's right - any mistakes, tell me and I'll fix it.

 

[ToO LaZy ^*]

looks pretty right to me...cheers!
EDIT: anyone for Q1?...i didn't really understand calculon's answer (prolly coz i'm slow)

 

[CrashOveride]

You can use Simpson's rule for volumes ? :S
Damn, ive only ever used it to find areas under curves

 

[ToO LaZy ^*]

that's what i was thinking....???????

 

[ToO LaZy ^*]

Originally posted by ToO LaZy ^*

EDIT: anyone for Q1?...i didn't really understand calculon's answer (prolly coz i'm slow)

i got it..it's allllll good