How do I calculate the volume of this?

[-EquinoX-]

How do I calculate the volume of this??

What is the volume of the region bounded by y = x^2, y = 1, and the y-axis rotated around the y -axis

 

[mathwonk]

define a voklume function, as a function of y, by letting V(y) be that portion of the volume lying below height y. then the derivative of this function i the area of the circular face of this portion of volume, i.e. dV/dy = πr^2 where r = x = sqrt(y), so dV/dy = πy. so you guess a formula for V(y) and then plug in y = 1.

 

[-EquinoX-]

I should do this using integral.. I was thinking of slicing this region vertically but then how do I represent this in integrals?? I have to take integrals from -1 to 1 right?

I would evaluate this as the integral from 0 to 1 of (1-squareroot of y)^2 dy

is that right??

 

[HallsofIvy]

Why in the world would you slice it vertically? Since it is rotated around the y-axis, slices horizontally will be disks with radius x. Each would have area $\pi x^2= \pi y$ and each infinitesmal disc will have volume $\pi y dy$. Integrate that.

 

[-EquinoX-]

Why in the world would you slice it vertically? Since it is rotated around the y-axis, slices horizontally will be disks with radius x. Each would have area $\pi x^2= \pi y$ and each infinitesmal disc will have volume $\pi y dy$. Integrate that.

I am sorry that's my mistake. I would slice it horizontally and take the integral from 0 to 1. And shouldn't the radius be 1-square root of y?? Because it's the intersection of y = x^2 and y = x

 

[HallsofIvy]

I am sorry that's my mistake. I would slice it horizontally and take the integral from 0 to 1. And shouldn't the radius be 1-square root of y?? Because it's the intersection of y = x^2 and y = x

Where did you get y= x from? Your original post was:

What is the volume of the region bounded by y = x^2, y = 1, and the y-axis rotated around the y -axis

Every horizontal "slice" is a circle with center at the y-axis, x= 0, and the end of a radius at $x= \sqrt{y}$. Of course, the area of the disk is $\pi x^2= \pi y$.