More volume of a washer cross section(integration)

  • Thread starter Thread starter Neutrinogun
  • Start date Start date
  • Tags Tags
    Cross Volume
Neutrinogun
Messages
7
Reaction score
0

Homework Statement



Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.


Homework Equations



y = x^2, x = y^2 about y = -1


The Attempt at a Solution



I tried using both cylindrical and washer methods - but for cylindrical I couldn't figure out what the shell radius would be, so I switched to washer.

\pi\int_0^1(x^2+1)^2 - (√x+1)^2 dx
\pi\int_0^1 x^4 + 2x^2 - x - 2√x dx
\pi(x^5/5 + (2x^3)/3 - x^2/2 - (4x^3/2))/3)
\pi(1/5 + 2/3 - 1/2 - 4/3)
-29\pi/30

And I get a negative volume. Any ideas?
EDIT: Ah. I just noticed I subtracted in the wrong order - would that fix it?
Edit2: Also, would you mind helping me with using the cylindrical method for this? Because I wasn't able to narrow down the shell radius - there was always a little unknown portion.
 
Last edited:
Physics news on Phys.org
Neutrinogun said:

Homework Statement



Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

Homework Equations



y = x^2, x = y^2 about y = -1

The Attempt at a Solution



I tried using both cylindrical and washer methods - but for cylindrical I couldn't figure out what the shell radius would be, so I switched to washer.

\pi\int_0^1(x^2+1)^2 - (√x+1)^2 dx
\pi\int_0^1 x^4 + 2x^2 - x - 2√x dx
\pi(x^5/5 + (2x^3)/3 - x^2/2 - (4x^3/2))/3)
\pi(1/5 + 2/3 - 1/2 - 4/3)
-29\pi/30

And I get a negative volume. Any ideas?
EDIT: Ah. I just noticed I subtracted in the wrong order - would that fix it?
Edit2: Also, would you mind helping me with using the cylindrical method for this? Because I wasn't able to narrow down the shell radius - there was always a little unknown portion.
For the cylindrical method, the radius is y - (-1), which is y + 1 .
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top