Volume of solids

[IsaaKC]

I'm having trouble coming to proper solution.

the problem is:
x=y^2, y=x-6, being revolved around the y-axis

I've made limits on the first integral, from 0 to -6, with the integral being (x+6)^2, multiplied by pi

I've made a second integral, (x+6)^2 - (y^2)^2, with limits 3 to 0, this also being multiplied by pi.

what I come with is ?(1/3y^3 + 6y^2 +36y) limits 0 to -6 plus ?(-1/5y^5 + 1/3y^3 + 6y^2 +36y) limits 3 to 0, but this is not the right answer.
where is my mistake?

 

[galactus]

It would appear your lower limit should be -2 instead of 0.

\(\displaystyle \pi\int_{-2}^{3}\left[(y+6)^{2}-y^{4}\right]dy\)

 

[IsaaKC]

How did you get that?

It's right, but I wanted to split it from 0 to -6 because that area forms a disk and from 3 to 0, there is a washer.

 

[galactus]

Here is a graph. The black horizontal line indicates a cross section.

The dashed lines represent the rotated image.

Your second integral is OK. You do not need two because the integration is w.r.t y

 

[IsaaKC]

I thought about that, but the problem I was having was y^2=x, so y=?x.

y=?x cannot exist below zero, that's why I splt it up.

 

[galactus]

Actually, \(\displaystyle x=y^{2}\Rightarrow y=\pm\sqrt{x}\)

 

[galactus]

But, we could use shells as well. This is where breaking it up would be a good idea.

\(\displaystyle 4\pi\int_{0}^{4}x\sqrt{x}dx+2\pi\int_{4}^{9}x(\sqrt{x}-(x-6))dx\)

Get the same answer. Though, it is more complicated than using washers.

 

[IsaaKC]

Oh right
That makes sense.

It had to be w.r.t. y, so I didn't try that method.

Thank you so much though!