problem (1 Viewer)

[doiyoubi]

I think one of the answers in cambridge is wrong for Furthur Questions 6 question 3, i get pi(ln2-1/2) but they get pi(ln2)
can someone confirm this with me thanks

 

[Raginsheep]

Can you post the question?

 

[doiyoubi]

find the volume by rotating the region
0<x<1, 0<y<1/(1+x²) about the y axis, using method of slicing

 

[doiyoubi]

and also this question

0<x<1, 0<y<e^-x2 about the y axis, using method of slicing

 

[NickP101]

I got PI ln2.

I used integral from .5 to 1 (draw the curve). Then added the volume of the cylinder under that curve.

 

[NickP101]

Sorry was a bit short before,

As the curve 1/(1 + x^2) is at 1 when x = 0 and at .5 when x = 1 you cant take the integral from 0 to 1 with respect to y. So take the integral from .5 to 1 and add the volume of the cylinder left, which r = 1, h = .5.

If you need anything more ill try write out the solution.

 

[doiyoubi]

Could you go into a bit more detail please?

 

[Riviet]

If you try to integrate betweent he limits y=0 and y=1, you are essentially trying to find the total bounded area between the curve and the y-axis, which fails because the curve approaches the asymptote y=0 and either end and so the rotated area is indefinite.

The region bounded in 0<y<1/2 and -1<x<1 (rotated about the y-axis) is essentially the rectangular region (rotated about the y-axis) that you're trying to find (as well as the top bit from 1/2<y<1). Therefore you need to split up the volume and basically do what nick said previously.