Ap Calculus HELP!

[Mo0o]

Ok, ive spent like and hour trying to figure out these two problems (similar)

Find the volume of the solid generated by revolving the region bounded by y=x^(1/2) and the lines y =2 x=0

a) about the line y=2
b)about the line x=4

please show me!, i have the answer but i need to know how to do it THANKS!

 

[Mo0o]

^bump^ cmon guys, help me out here

 

[MrBond]

It requires integration. There's a couple ways to do it, the shell method or the slice method (thats how I was taught anyway).

Unforuntatly, I can't help you more. I haven't done volumes in a long time, and any notes I have on it are quite a distance from me right now

 

[Mo0o]

yeah, i can do it about the two axises, but i cant do it to other lines

 

[Ns1]

well, what are you having problems with?

and are you using disks/washers or cylindrical(sp?) shells?


well with the lines all it will be is eitehr x +- 2/4 (respectively)

it's up to you to think it out (it's really easy to do if you draw it out)

 

[MrBond]

<< it's up to you to think it out (it's really easy to do if you draw it out) >>

Indeed. I've solved many a difficult problem in calculus by dawing a picture.

I had trouble with volumes at first too, thought they were a piece of cake when we got to multivariable calc

 

[Mo0o]

im having trouble with general setup

so for a) i should i have it integrate from 0 to 2 of (x^(1/2)+2)^2 then multiply by pi?

 

[Random Variable]

TRIPLE EDIT: The correct answers are in my other post.

 

[PowerEngineer]

Well...let's see what I remember. The Y=X^(1/2) line crosses X=0 at 0, and crosses Y=2 at X=4, making the cross section the area enclosed by moving from 0,0 vertically to 0,2 then horizontally over to 4,2 and then back to 0,0 along the Y=X^(1/2) line. The easiest way to find the area is to subtract the integral (from X=0 to X=4) of X^(1/2) from the are of the box (4*2=8). I think the integral of X^(1/2) is 0.5*X^(3/2), which means the area is 0.5*(4)^(3/2) - 0 = 4, which surprises me some because this means the cross sectional area is also 4. Now this cross section needs to be multiplied by the area of the circle (or donut) created by revolving it around the two lines. For A, the radius is 2 (from y=2 to y=0) making the area 4*pi, so the volume should be 16*pi. Similarly, B is 64*pi. At least, that's how I think it works -- chances are good I've made at least one error. Good luck.

 

[alee25]

<< Well...let's see what I remember. The Y=X^(1/2) line crosses X=0 at 0, and crosses Y=2 at X=4, making the cross section the area enclosed by moving from 0,0 vertically to 0,2 then horizontally over to 4,2 and then back to 0,0 along the Y=X^(1/2) line. The easiest way to find the area is to subtract the integral (from X=0 to X=4) of X^(1/2) from the are of the box (4*2=8). I think the integral of X^(1/2) is 0.5*X^(3/2), which means the area is 0.5*(4)^(3/2) - 0 = 4, which surprises me some because this means the cross sectional area is also 4. Now this cross section needs to be multiplied by the area of the circle (or donut) created by revolving it around the two lines. For A, the radius is 2 (from y=2 to y=0) making the area 4*pi, so the volume should be 16*pi. Similarly, B is 64*pi. At least, that's how I think it works -- chances are good I've made at least one error. Good luck. >>




i think i see your mistake. The intergral of x^.5 would actually be the ANTiderivitive of that, not the derivitive. So it makes it 2/3 X^(1.5), and when you find the intergral: 2/3 (4^1.5) - 0 = 16/3. I dont see why you subtracted your 8-4 though, becasue you already found the area of the shape by using the intergral

Edit: does the volume equal the cross section multiplied by the top circle?

 

[PowerEngineer]

Ah well...I wasn't sure about the multiplier. Anyway, the area I was caluclating is actually outside the cross section. The cross section is the area of the rectangle minus the area under the curve.

The volume should be the area of the cross section multiplied by the area of the circle or donut formed by the rotation (I think).

 

[alee25]

oh i see why you subtracted 8-4, i was thinking about the second part (B).

and then shouldn't (for part A) you multiply ((8 - (16/3)) = 8/3 x 16 pi because you use the x axis as the radius?

and for the second part B, you multiply 16/3 x 4pi because you use the y axis as the radius?

At least thats how it looks on my drawing.

So i get the answers: A. ((128PI)/3), B. ((64PI)/3)

 

[Mo0o]

according to the book the answers are

8pi/3 and 224pi/15

 

[Random Variable]

a) Integral pi*((2-(x^(1/2))^2 dx from 0 to 4 = 8*pi/3

b) Integral pi*((y^2-(4-y^2))^2 dy from 0 to 2 = 224*pi/15

 

[alee25]

vespian how did you get that?

 

[Martin]

ok, this sounds familiar, but WTF is a volume?

You mean the area under the curve? THe one you can find using Reimann Sums?

 

[Random Variable]

<< vespian how did you get that? >>



a) Graph y=x^(1/2), y=2, and x=0 (which is the y-axis), and then you'll notice that the radius of the bounded region is 2-x^(1/2)

b) Again, graph the two functions and x=0 and you'll notice that the outside radius of the bounded region is y^2, and that the inside radius of the region is 4-y^2; therefore, you have to use the washer method. And because the solid is being rotated in the horiztonal direction, you have to have everything in terms of x.