Why this solids also lives below z axis?

Question

The base of a certain solid is the circle $x^2 + y^2 = a^2$.
Each plane perpendicular to the x-axis intersects the solid in a square cross-section with one side in the base of the solid.
Find its volume.

Here is my try :
$$y = \sqrt{a^2-x^2}$$
$$dV = (a^2-x^2)dx$$
$$V= 8\int_0^a (a^2-x^2)dx = \frac{16}{3}a^3$$
which is the correct answer given by textbook.

I know that solid has eight octants, for the book answer is $16a^3/3$,
Can someone help me understand why it has 8 octants, I integrate one part of the solid and then multiply by eight, why do I have to multiply by eight instead of four?

If the solids live above the z-axis then I have to multiply by four but in this case, how do I know the solid is also above the z-axis then multiply by eight

Answer 1

For the square cross-section at a particular $x$, the side length on the base is $2\sqrt{a^2-x^2}$, so the cross-sectional area should be $4\left(a^2-x^2\right)$ -- $4\times$ what you used for $dV$.

$$dV = 4\left(a^2-x^2\right)dx$$

As the circular base goes from $x=-a$ to $x=a$, the limits of integration may instead be from $-a$ to $a$. This explains $2\times$ out of the scaling factor in your $V$.

These together explain the seemingly arbitrary $8\times$ scaling factor in your $V$.

$$\begin{align*} V &= \int_{-a}^a 4\left(a^2-x^2\right) dx\\ &= \int_{-a}^0 4\left(a^2-x^2\right) dx + \int_0^a 4\left(a^2-x^2\right) dx\\ &= \int_{a}^0 4\left(a^2-(-x)^2\right) d(-x) + \int_0^a 4\left(a^2-x^2\right) dx\\ &= \int_0^a 4\left(a^2-x^2\right) dx + \int_0^a 4\left(a^2-x^2\right) dx\\ &= 8\int_0^a \left(a^2-x^2\right) dx\\ &= 8\left[a^2x - \frac{x^3}{3}\right]_0^a\\ &= \frac{16}{3}a^3 \end{align*}$$

(Also, in an earlier version, you had an incorrect $\pi$ factor that got lost after incorrect integration.)

Answer 2

Just like the $x-y$ plane has four quadrants,

• $+x, +y$
• $+x, -y$
• $-x, +y$
• $-x, -y$

the $x-y-z$ space has eight octants:

• $+x, +y, +z$
• $+x, +y, -z$
• $+x, -y, +z$
• $+x, -y, -z$
• $-x, +y, +z$
• $-x, +y, -z$
• $-x, -y, +z$
• $-x, -y, -z$

Answer 3

The answer is straightforward. Coordinates $x$ and $y$ will only have positive or negative signs. That means they can have up to 4 variations, the same as in quadrants. Therefore, if we put another coordinate $z$, number of possibilities increases to 8, as $2^3 = 8$

Possibilities are:

• $+x,+y,+z$
• $+x,+y,−z$
• $+x,−y,+z$
• $+x,−y,−z$
• $−x,+y,+z$
• $−x,+y,−z$
• $−x,−y,+z$
• $−x,−y,−z$

Answer 4

The Question wording requires Multiplication by $4$ , not by $8$ . . . .

[[ EITHER the OP Integration is wrong (by the factor of $1/2$) & the wrong Multiplication by $8$ rather than $4$ somehow gives the right Answer (by giving the Extra factor $2$ thus cancelling the factor of $1/2$) OR the textbook may have a typo or error : Let us see which ]]

Earlier , My Answer was targeted towards whether (1) the Solid was in 4 Octants or 8 Octants & (2) whether to Multiply 1 Octant Volume by 4 or to Multiply 1 Octant Volume by 8 , assuming the OP did the Correct Integration.
I have updated my Answer to cover that Integration Itself , to verify what is going wrong where !

Here is the Image :

The Black Circle in $XY$ Plane is the BASE of the Solid.
Every Plane Perpendicular to $X$ Axis will get a Square Cross-Section (Shown in Green here) & we have to integrate that Square Area.
The TOP PART of the Solid (Shown in Blue) makes a Semi-Circle on the $XZ$ Plane. The Blue Curve will not be a Whole Circle , because the Black Circle is the BASE.
Thus we calculate in ONE Quadrant (ONE Octant) in $XY$ Plane & then Multiply that by $4$.

Imagine the Purple line going through the Solid & Parallel to the $YZ$ Plane & going through the $X$ Axis. The Solid Exists when $z$ Value is between $0$ & some Positive Value. There is no Solid when $z$ Value is Negative.

If the Black Circle was not the BASE but the Interior , then we might have a Different Integral with Different Criteria & we might have to multiply by $8$.

Integration :

The Square at Distance $x$ along the $X$ Axis has BASE Corners at $(x,+\pm \sqrt{a^2-x^2})$ , which gives width of the Square $2\sqrt{a^2-x^2}$ which is also the height of the Square.
Now , the Square is half in 1 Octant & half in other Octant. Thus Area in 1 Octant is $height \times width/2$ which is $2\sqrt{a^2-x^2} \times 2\sqrt{a^2-x^2}/2 = 2(a^2-x^2)$

We want to Integrate this in 1 Octant , thus $x$ is between $0$ & $a$.
Volume in 1 Octant is $\int_0^a 2(a^2-x^2) dx = [a^2x-x^3/3]_0^a = a^3-a^3/3=2a^3/3$
Volume in 4 Octants is $4 \times 2a^3/3 = 8a^3/3$

Thus we see that OP had 2 errors , which got cancelled to get the Correct Answer , by luck !

SUMMARY :

Assuming the Question wording is Accurate : You are right , you have to multiply by $4$ , but your Integration was having Extra factor $1/2$ & when you multiply by $8$ , it gives Extra factor $2$ which cancels the $1/2$ to get the text book Answer by luck.
In other words , $4 \times V \equiv 8 \times V/2$ & Everything is OK !