1. ## Integral Volume

I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
Given that Volume = integral between a and b of A(x) dx so I came up with:

Circle: 4 (integral between -4 and 4) (16-x^2) dx
Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
Semi: (integral between -4 and 4) (pi/2) (16-x^2)
Isoc: (integral between -4 and 4) (16-x^2)

But those aren't working out?

2. Originally Posted by BooGTS

I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
Given that Volume = integral between a and b of A(x) dx so I came up with:

Circle: 4 (integral between -4 and 4) (16-x^2) dx
Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
Semi: (integral between -4 and 4) (pi/2) (16-x^2)
Isoc: (integral between -4 and 4) (16-x^2)

But those aren't working out?
Square.

3. Where I say "Circle" It should have read square and I apologize for the pictures not loading in the correct order for my equations.

4. Originally Posted by BooGTS
Where I say "Circle" It should have read square and I apologize for the pictures not loading in the correct order for my equations.
So no circle then?

5. Originally Posted by BooGTS

I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
Given that Volume = integral between a and b of A(x) dx so I came up with:

Square: 4 (integral between -4 and 4) (16-x^2) dx
Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
Semi: (integral between -4 and 4) (pi/2) (16-x^2)
Isoc: (integral between -4 and 4) (16-x^2)

But those aren't working out?
Originally Posted by dwsmith
So no circle then?
It should read like in the quote above. The circle on the bottom is still x^2 + y^2 = 16, but the cross section is a square, eq triangle, isoc triangle, and semicircle. Sorry for the confusion I created!

6. Originally Posted by BooGTS
It should read like in the quote above. The circle on the bottom is still x^2 + y^2 = 16, but the cross section is a square, eq triangle, isoc triangle, and semicircle. Sorry for the confusion I created!
Area of a squate is $A=s^2$

$x=\sqrt{16-y^2}$

$s=2\sqrt{16-y^2}$

$s^2=4(16-y^2)=A(y)$

$\displaystyle\int_a^b A(y)dy \ \text{or} \ 2\int_0^b A(y)dy$

7. Originally Posted by BooGTS

I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
Given that Volume = integral between a and b of A(x) dx so I came up with:
With $x^2+ y^2= 15$, $y= \pm\sqrt{16- x^2}$ so for given x the base in each figure is $2\sqrt{16- x^2}$

Circle (but apparently you meant square): 4 (integral between -4 and 4) (16- x^2)dx
The area of a square of base s is $s^2$. Here, $s= 2\sqrt{16- x^2}$ and its square is $4(16- x^2)$.
Yes, that looks like exactly what you have.

Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
An equilateral triangle with base s has height $\frac{\sqrt{3}}{2}s$ and so area $\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)\left(2\ sqrt{16- x^2}\right)\left(2\sqrt{16- x^2}\right)$.
Looks to me like you are missing a factor of 2.

Semi: (integral between -4 and 4) (pi/2) (16-x^2)
The area of a semi-circle with diameter s is $\frac{\pi}{4}s^2$ so with $s= 2\sqrt{16- x^2}$, the area will be $\pi(16- x^2)$
Again, you are a factor of 2 off.

Isoc: (integral between -4 and 4) (16-x^2)
The height of an isosceles triangle is independent of the base.
You can't find the volume without knowing what the height is supposed to be.

But those aren't working out?