trigonometry please help (1 Viewer)

[AFGHAN22]

for the curves y=sinx and y=cosx for 0 /<x/< pi/2 determine the magnitude of the area bounded by the y-axis and the curves y=sinx, y=cosx
calculate the volume of the solid obtained when this area is rotated about ther x -axis.

 

[zeek]

use integration

 

[_ShiFTy_]

With y=sinx and y=cosx around the x axis, just use the 3U formula
When its around the y axis, you'll have to use by parts when integrating (i think)

 

[AFGHAN22]

that doesn't help

 

[zeek]

what do u mean by 0 /< pi/2

 

[AFGHAN22]

sorry its meant to be 0 /< x /< pi/2

 

[Riviet]

y=sinx and y=cosx intersect at π/4, so I think the area the question is referring to is given by
π/4
∫ cosx - sinx dx
0

Then volume should be

π/4
π∫ cos²x - sin²x dx
0

This second integral is easy if you use a double-angle formula.

 

[hyparzero]

Let sin(x) = cos(x)
Therefore
tan(x) = 1

Hence x = pi/4

A = ∫[cos(x) - sin(x)]dx between x = 0 and x = pi/4

A = [sin(x) + cos(x)] {pi/4 ~ 0}

A = (1/Rt(2) + 1/Rt(2)) - (1)

Therefore

A = Rt(2) - 1

For volume, use whatever method you've learnt, like slicing, or cylindrical shells. etc...