help with some calc 2

[mecbluefugate]

So I'm reviewing for an exam that I have tomorrow, and I came across these two similar problems. I realized I completely forgot how to do them. I know they both have to do with the integral of pi R squared, and something to do with outter radius - inner radius, but I am drawing a blank on how exactly to do them. Probably due to cramming, but somebody help me out with these please ---

1. Find the volume of the solid generated by rotating the region bounded by the curves y=xx and x=yy about the y-axis.

2. Find the volume of the solid generated by rotating the region bounded by the curves y=3x-xx and y=0 about the line x= -1.

Again, just trying to figure out how to do these problems. Since I'm reviewing for an exam, the process is much more important than the raw answer.

[Drag]

I'll explain how to do the following problem:
Find the volume of the solid generated by rotating the region bounded by y=f(x), for x belonging to [a, b], around the x axis.
(Your problems are the same, you just specify f(x), move the axis of rotation, etc.)

A small region of [x, x+dx] will generate a disc with the thickness dx and the radius of R=f(x), its voulume dV=pi*R^2*dx. The whole volume of the body will be the summ of all such small volumes (I omit the rigorous approximation and taking the limit approach), or in the limit of dx going to 0 an integral:
V= Summ of dV = Intgeral (from a to b) { pi*f(x)*f(x)*dx }

[jay_shark]

h(x)=x^2 and f(x)=+-sqrtx and notice that h(x) and f(x) intersect at (0,0) and (1,1)

g(x)= h(x)-f(x)
g(x)= x^2-sqrtx

So we want the integral(or area under the curve) of g(x) evaluated at x=0 and x=1 .

G(x) = 1/3*x^3 -2/3*x^(3/2) | x=0 and x=1
G(1) - G(0) = -1/3 - 0 = -1/3

So the area bounded by the two equations is 1/3 since it cannot be a negative number .

2) Use this formula :

V= pi* integral of f(x)^2dx |x=a to x=b

V=pi* integral of (3x-x^2)^2dx |x=0 to x=3

To solve for a and b I solved the equation 3x-x^2 =0
x(3-x)=0 , x=0 or x=3 .

V=pi* int [9x^2 -6x^3 + x^4)dx | x=0 to x=3
V=pi*[3x^3 - 6/4*x^4 + 1/5*x^5] |x=0 to x=3

Now just solve for V(3) - V(1) and we're done .

[mecbluefugate]

Thanks, figured it out at around 5 AM this morning.

[relativity_x]

jay shark did this incorrectly, and the fact that he obtained -1/3 as a volume should have made it obvious.

To solve the problem, find the cross sectional area and integrate to get volume.

If you use shell's method for 1, you get the height of the solid is (sqrt(x)-x^2). You still need the radius which is equal to sqrt(x).

This makes the volume=2*pi*integral(sqrt(x)*(sqrt(x)-x^2)) evaulated from 0 to 1.

volume=2*pi*[1/2*x^2-2/7*x^(7/2)] evaluated at x=0 to x=1

After doing the algebra, you obtain V=3/7*pi.

2)You can also use shell's method here. Except the axis of revolution is translated to the left by 1, so radius would be (1+x). You should be able to find the height. From there, you set up the integral and evaulate it from 0 to 3.

[jay_shark]

[ QUOTE ]
jay shark did this incorrectly, and the fact that he obtained -1/3 as a volume should have made it obvious.



[/ QUOTE ]

Relativity from pocket 5's ? It must be you .

Anyway , I already gave the formula for solving these types of problems in my second solution . When I first read the first problem I thought it was for AREA and not volume which is why my answer was 1/3 . The fact that I got a negative number has nothing to do with my calculation being wrong . Just insert absolute values to the number to correct the problem .

Are you doubting that the area of the bounded regions is 1/3 ?

[relativity_x]

your second one is wrong too.

V=2*pi*integral((x+1)*(3x-x^2)) from x=0 to x=3.

[jay_shark]

You do realize that there is more than one way of getting your answer , right?

Try googling the Volume of Revolution formula and you should get V=pi*{int of [f(x)]^2dx |x=a to x=b}

[relativity_x]

yes, i realize there are other ways to do it, but your expression is not equivalent to mine.

[jay_shark]

Question 2 should have been stated that we're rotating about the line x=0 or the y-axis .

Since y=0 is under the curve y=3x-x^2 from (0,3) but above the curve from (-1,0)

[relativity_x]

no it shouldnt. You can revolve any solid around any arbitrary axis. Generally, the x or y axes are chosen. In this situation, x=-1 was selected as the axis; therefore, you have to account for that with the radius (x+1).