Volumes of solids cross sections

[patent umbra]

I was trying to find the volume of this solids cross section and i kept getting a negative volume. I have attached my work but i dont see why I am getting a negative volume for the bottom 2 circles

[olive cedar]

Your expansions of the brackets^2 are wrong

$(a-b)^2 = (a-b)(a-b) = a^2 - 2ab + b^2$

[runic silo]

TayBee

[olive cedar]

Take a look at your (6-y)^2 and (2-y)^2 and expand then integrate again

[patent umbra]

i dont understand

[olive cedar]

My bad, you simplified at the same time and tbh didn't bother checking what happened after bringing the 4s outside the bracket in lmao

Will have another look, one sec

[olive cedar]

Right @patent umbra next things I've spotted having actually run through the maths rather than just taken a cursory glance XD

You have your circles the wrong way around

As in, you've said your bottom circle is defined by x^2 + (6-y)^2 = 4 and thus integrated that from 0 to 4

That's your middle circle (you can see that because you know how the equation of a circle is defined so it's going to have centre (0,6) or you can just check your desmos output)

So you should be integrating that from 4 to 8

And similarly the circle defined by x^2 + (2-y)^2 = 4 should be integrated from 0 to 4

Incidentally, I also don't understand why you've chosen your constant outside the integral as pi/2 and not pi? Your solid of revolution is given by $$\pi \int y^2 dx$$

[runic silo]

TayBee

[olive cedar]

When you revolve the circle around the y axis it is going to generate a sphere so unless you've been told to revolve it 90 degrees (or the semicircle by 180 degrees) there's no need to halve it, which brings me to my next point

You can quite readily avoid this suffering by considering the geometry of this problem

As in, the solid of revolution of a (semi)circle is a sphere

And we know the volume of a sphere is 4/3 pi r^3

Radius here of both is 2

So each circle has a solid of revolution of 4/3 pi * 8 = 32/3 pi

SImilarly, the solid of rev of a rectangle (/square) is a cylinder, volume pi r^2 h

You have a cylinder height 1 and radius 0.5

And second point: even if you want to go down the calculus route, the circles are identical, so you can just multiply the result from the bottom circle by 2

You can also 'bring down' each shape to touch the x axis and integrate them with your lower limit as 0 and reduce the upper limit accordingly to make for easier limit substitution as you can see that it won't change the volume

[patent umbra]

hum'

well from what you said

sorry there is alot of it

but this is what i came up with now

[patent umbra]

that is that part fixed

i guess im just confused on my cylinder up top still

im not sure if that is the right equation

[olive cedar]

Much better, although as I said your volumes are still half of what I expect them to be because of this

Since you're revolving around the y-axis, your volume is given by $$\pi \int x^2 dy$$

[runic silo]

TayBee

[olive cedar]

For your square at the top that revolves to make a cylinder, using calculus, we know the line we're revolving is x=1/2

$\pi \int_{8}^{9} \left( \frac{1}{2} \right)^2 dy$

[runic silo]

TayBee

[patent umbra]

well that is because my cross sections are semi circles.

this was my thinking for it

[olive cedar]

Two ways of thinking about it, either you're revolving a semi circle through 360 degrees, in which case the formula I wrote holds, or you're revolving the entire circle through 180 degrees, in which case you half it because it's 180 degrees but then double it again because of the symmetry of the function

tl;dr regardless, it remains as pi rather than pi/2

As I say, you can (and should where you can) confirm the values achieved from calculus using what we know from geometry, because you know how to calculate the volume of spheres and cylinders which is what the solids of revolution of these functions are

Those are your standard formulae, worth committing to memory (but by the time you've done it 8,000 more times in practice questions, you'll have them memorised either way lmao)

[patent umbra]

yes i agree but instead of revolving these are cross sections of solids. and our cross section for this solid is semi circles which is why i have put the pi/2 on the outside of every integral

as in this

taking the area of a circle and changing it to volume of cross sectinos

[patent umbra]

also what would my x value b? i assume it would just be .5 (a constant) for every spot

[olive cedar]

Can you screenshot the original question, I'm probably just being stupid but I don't rly understand what you mean by this, like what solid are you creating here if not a sphere

[olive cedar]

Yeah you're right, x=1/2

[patent umbra]

there is no question but there is a rubric, that is the closest i have

[olive cedar]

Oh I see, so what we're looking at in the 2D representation is the base which semi-circles will be perpendicular to? So it's like 2 hemispheres/domes and a half cylinder/tunnel

In which case yes, now you've fixed your limits, your answers are correct

Assuming the cross sections for all 3 objects are semi-circles

[patent umbra]

Yes but I'm just confused on the cylinder up top

On how to solve for thr x value

Like how do i solve for my volume on seperate pieces, for example, what is my volume at y= 8.25?

i assume i would have to take my x/r value and solve for something but im not sure

[olive cedar]

Nah your current method is correct, your semicircle has area pi/2 * r^2

Your radius is just a constant value

x=0.5

Nothing further required

So at every single y value in your limits, r = 0.5, thus 2pi*r^2 is the same

It's exactly the same as your circles where you made x the subject and then integrated

Just now we don't need to do that because your equation is already in x

[patent umbra]

Nono like I need to make my physical model too as mentioned in this rubric

To do that I need to know the volume at different points such as

8.1875
8.375
..

.