cssa trial (1 Viewer)

[live.fast]

the area between the curve y = (sinx) ^2 and the x axis between x = 0 and x = (pi / 2) is rotated through one complete revolutio bout the x axis

(i) find the exact value of the area of the region

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

 

[ellen.louise]

the area between the curve y = (sinx) ^2 and the x axis between x = 0 and x = (pi / 2) is rotated through one complete revolutio bout the x axis

(i) find the exact value of the area of the region

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

Remember this result:

cos2x = 1 - 2sinxsinx
sinxsinx = 1/2(1-cos2x)

integral sign = }

i)
A = } pi/2 , 0 , sinxsinx . dx
= } pi/2 , 0 , 1/2(1-cos2x) . dx
= 1/2 } pi/2 , 0 , (1-cos2x) . dx
= 1/2 [x - 1/2sin2x] , pi/2 , 0
= 1/2 [(pi/2 - 1/2sinpi) - (0 - 1/2sin0)]
= 1/2 [pi/2 - 0]
= pi/4 units^2

ii)
fuck i can't be bothered right now...

 

[ianc]

yeah you must be able to integrate sin²x and cos²x because they will ALWAYS appear in an extension one exam.....


although the Simpson's rule one should be easy if you just give it a little thought....

remember that the volume formula is π∫y² dx.....(its derived from the volume of a cylinder, which is πr²h)

So apply Simpson's rule to [sin²x]² on the interval 0 to π/2, then multiply the result by π to get your volume in units³