stuck :( (1 Viewer)

[Mc_Meaney]

OK, Im having trouble with the following (see attached diagram also)
Calculate in terms of pi

(I) The shaded region of PQRS

(II) The perimeter of the shaded region PQRS


Apologies for a shonkie diagram, my original one was 10 times to big to upload

 

[Mc_Meaney]

Heres the working Ive got...dunno if its good (obviously lol)(I) 1/6.pi.(25)^2 - 1/6.pi.(15)^2= 104 1/6pi - 37.5 pi= 66.6....picm^2(1dp)(II) QR = 26.2 PS = 15.7 PQ + RS = 20Add and get 61.9cmAhhh I hoep thats right...

 

[Mountain.Dew]

yeah they seem okay to me.

just one small thing though: with (ii), please leave ur answers in terms of pi. please dont round off.

 

[Mc_Meaney]

yeah they seem okay to me.

just one small thing though: with (ii), please leave ur answers in terms of pi. please dont round off.

Oopps. THat is the length though....so just have it 61.9pi cm?

 

[Mc_Meaney]

THanx moundain dew

 

[Mc_Meaney]

Find the area bounded by y = x/(x^2+1), x-axis and the lines x=2 and x=4 to 2dp.

Help :S

 

[Mc_Meaney]

Area between y=Inx, y=2 and y=4

 

[Riviet]

For the first, observe that x/(x²+1) = 1/2. 2x/(x²+1), which has an expression in the denominator with its derivative in the numerator.

Isolate x for the second one:

y=lnx
x=e^y

Using the boundaries y=2 and y=4, you can now integrate this expression with respect to y.

Hope that helps.