- #1

- 564

- 1

**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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter -EquinoX-
- Start date

- #1

- 564

- 1

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

- #2

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,113

- 1,317

- #3

- 564

- 1

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??

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

is that right??

Last edited:

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

- #5

- 564

- 1

horizontallywill be disks with radius x. Each would have area [itex]\pi x^2= \pi y[/itex] and each infinitesmal disc will have volume [itex]\pi y dy[/itex]. 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

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

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

Every horizontal "slice" is a circle with center at the y-axis, x= 0, and the end of a radius at [itex]x= \sqrt{y}[/itex]. Of course, the area of the disk is [itex]\pi x^2= \pi y[/itex].What is the volume of the region bounded by y = x^2, y = 1, and the y-axis rotated around the y -axis

Share: