Working out problem!!!!!! (1 Viewer)

[RHINO7]

I can't get the working out for these questions correct!


i) Find the area between the curve y= sec^2 x/4, the x-axis and the lines x= pi/4 and x= pi/2, correct to 2 decimal places.


ii) Find the volume, correct to 2 decimal places, of the solid formed when the curve y=sec pi x is rotated about the x-axis from x=0 to x=pi/6.


For some reason I can't work out the sec^2 questions.

 

[RHINO7]

Please!

 

[jb_nc]

What are you having trouble with?

1) The intergral of sec^2 x is tan x. Upper and lower bounds are given in the question. You should be able to do it.

2) Think to yourself, "square the opposite of what it is rotated around". So around x, square y. You should know you need sec^2 x anyway because 2 unit techniques don't teach you how to integrate sec x. Upper and lower bounds are obvious in this case. Don't forget to multiply the intergral by pi, V = pi (integral from a to b) y^2 dx.

Post your working and I'll tell you if you're right or not.

 

[RHINO7]

i)A={ 1/1/2 tan x/4

= [ 2 tan x/4]pi/2 , pi/4

[2 x pi/2/4]-[2 x pi/4/4]


ii)y^2=sec^2 pi x

V= pi[1/pi tanpix] pi/6, 0

pi[tan pipi/6]-[tanpi(0)]

 

[RHINO7]

Yes. Have I done anything incorrect.