Volumes and Trig Functions (1 Viewer)

[Avenger6]

I'm wondering if anyone can help me with the following questions:

In question 5 i get 0 as answer (which is obviously wrong), I think it may have something to do with my integration as when I integrate y=sec^2(x/4) i get y=4tan(x/4)...is this correct? Im lost all together on question 6 lol...i've integrated y=sec(pi)x to get y=1/(pi)tan(pi)x which i don't think is correct.

Any help is greatly appreciated.

 

[tommykins]

5) int sec^2 x/4 = 4tan x/4

subbing in pi/4 and pi/2

You get 4tan pi/8 - 4 tan pi/16 = 1.6568... - 0.79564... = 0.86 units^2

6) y = sec pix, but since we're finding volume, it is int y^2 dx thus, y^2 = sec^2 pix

integration now becomes V = pi int sec^2 pi x dx from pi/6 -> 0

V = pi int sec^2 pix dx pi/6->0
V = pi [ tan pix/pi] pi/6->0

Sub in the values and bingo

 

[Avenger6]

My mistake...for question 6 it should read "from x=0 to x=0.15". Sorry about that. (Now fixed in original post)

Im following everything up until this point:

V = pi int sec^2 pix dx pi/6->0
V = pi [ tan pix/pi] pi/6->0

When you integrate sec^2 pix shouldn't it become 1/pi tan pix dx becuase int sec^2 (ax+b)dx= 1/a tan (ax+b)???

 

[Avenger6]

Nevermind I worked it out. The integral was 1/pi tan pix but I forgot to put the pi in radians...so stupid. Thanks for the help.

 

[tommykins]

My mistake...for question 6 it should read "from x=0 to x=0.15". Sorry about that. (Now fixed in original post)

Im following everything up until this point:

V = pi int sec^2 pix dx pi/6->0
V = pi [ tan pix/pi] pi/6->0

When you integrate sec^2 pix shouldn't it become 1/pi tan pix dx becuase int sec^2 (ax+b)dx= 1/a tan (ax+b)???

Sorry, I should've made it (tanpix)/pi.