Volumes of solids of revolutions

[John_Smith01]

Homework Statement

In the question your working with the region bounded by the two curves y=x^2 and y=x^3.

In the first part of the question you had to revolve the region around the x-axis and find the area which I managed to do by subtracting two areas from each other.

The second part of the question asked to revolve the region around the y-axis which I didn't know how to do since I couldn't find (x^3)dy

 

[rock.freak667]

Revolving around the y-axis would give:

V= π∫ x² dy

so if y=x³, then x = y ^1/3 and hence x² = y^2/3

 

[John_Smith01]

That's what I tried before but kept getting the answer wrong. I think my textbook is wrong, did you get 1/10pi u^3?

 

[SammyS]

Revolving around the y-axis would give:

V= π∫ x² dy

so if y=x³, then x = y ^1/3 and hence x² = y^2/3

Using the washer method, the volume would be: V = π∫ ((x₂)²- (x₁)²)dy,

where x₂=y^(1/3) and x₁ = y^(1/2).

 

[rock.freak667]

That's what I tried before but kept getting the answer wrong. I think my textbook is wrong, did you get 1/10pi u^3?

I do

Using the washer method, the volume would be: V = π∫ ((x₂)²- (x₁)²)dy,

where x₂=y^(1/3) and x₁ = y^(1/2).

It will end up as the same since OP will have to subtract one from the other while that way will make into one simple integral.

 

[SammyS]

...
It will end up as the same since OP will have to subtract one from the other while that way will make into one simple integral.

When OP mentioned "subtracting two areas from each other", I thought he literally meant areas, not volumes. Oh well !