Calculus II: Vol. of Rev. about y-axis (region bounded by y^2 = x, x = 2y)

[reconrusty]

I have a Volume of revolution about the y-axis that just doesn't seem to be working for me.

y^2 = x is one equation and x = 2y is the other.

First i changed the y^2 = x to x = sqrt(y)

I graphed the two and figure out that x = sqrt(y) is the outer radius, and the x = 2y is the inner radius, with the interval being between 0 and 1/4.

I setup the integral as:

V = pi (integral sign) (sqrt(y)^2 - (2y)^2) dy, with the lower bound being 0 and the upper being 1/4.

I evaluate the integral and I end up with 0.5y^2 - 4/3(y^3), then plug in the values of 1/4 and 0, but it isnt giving me the right answer.

Any help??

 

[tkhunny]

I have a Volume of revolution about the y-axis that just doesn't seem to be working for me.

y^2 = x is one equation and x = 2y is the other.

First i changed the y^2 = x to x = sqrt(y)

I graphed the two and figure out that x = sqrt(y) is the outer radius, and the x = 2y is the inner radius, with the interval being between 0 and 1/4.

I setup the integral as:

V = pi (integral sign) (sqrt(y)^2 - (2y)^2) dy, with the lower bound being 0 and the upper being 1/4.

I evaluate the integral and I end up with 0.5y^2 - 4/3(y^3), then plug in the values of 1/4 and 0, but it isnt giving me the right answer.

Any help??

1) x = 2y and x = y^2 intersect at (0,0) and (4,2). Let's work on that, first.

2) Looking from the y-axis, maybe standing at (0,1), and moving in the direction of the positive x-axis, which do you encounter first? Let's figure out which is inner and which is outer.

3) You seem to be wanting to integrate along the x-axis. This may be why you solved for y = sqrt(x) and then became confused about the variables and changed it, somehow, to x = sqrt(y) -- Which is not particularly related to x = y^2. Don't do that. Integrating along the y-axis is just fine.

Okay, give it another go.

 

[Dr.Peterson]

I have a Volume of revolution about the y-axis that just doesn't seem to be working for me.

y^2 = x is one equation and x = 2y is the other.

So, you are revolving the region bounded by these curves about the y-axis? And since both are defined by x as a function of y, the natural way is to uses "washers", integrating with respect to y? That's all fine.

First i changed the y^2 = x to x = sqrt(y)

There's the main problem. How are x = y^2 and x = sqrt(y) the same thing? If anything, it would be y = sqrt(x), wouldn't it?

Don't change the equation; leave both as functions of y. You can graph them as they are, by treating y as the independent variable.

I graphed the two and figure out that x = sqrt(y) is the outer radius, and the x = 2y is the inner radius, with the interval being between 0 and 1/4.

I setup the integral as:

V = pi (integral sign) (sqrt(y)^2 - (2y)^2) dy, with the lower bound being 0 and the upper being 1/4.

I evaluate the integral and I end up with 0.5y^2 - 4/3(y^3), then plug in the values of 1/4 and 0, but it isnt giving me the right answer.

If you keep the equations as given, you will get the correct intersection point, and therefore the correct limits of integration, and the correct integrand.

Try that, and let us know what you get.