# Find the volume for example of a sphere, we slice it into numerous disks with small depth dr and get the infinitesimal volume dV.

Brenton Dixon 2022-07-23 Answered
Finding volume via integration
When we are asked to find the volume for example of a sphere, we slice it into numerous disks with small depth dr and get the infinitesimal volume dV. And then we sum those slices together by putting an integral symbol. But my question is the basic integration we learnt was to find area under the curve.In that case we proved that the area is simply the anti derivative of the function. But here how do we use integration by putting an integral symbol before Adr without proving that the derivative of the volume is the area of the disks? I am so disturbed by this as I didn't see any author giving the proof with proper intuition.
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## Answers (2)

salumeqi
Answered 2022-07-24 Author has 15 answers
Step 1
What the Fundamental Theorem of Calculus actually tells us is that in general, .
Applying this to your example gives us the required formula.
This is how we can apply it volume. Suppose we have a function f(x), and we want to revolve around the x axis the area bounded by $x=a$, $x=b$, the x axis and $y=f\left(x\right)$ to form a volume, V.
Consider a very small strip of the volume that is created, at a point x. This will be a cylinder with width $\delta x$ and its radius of its 2 flat surfaces will be f(x). So the volume of this cylinder is $\pi {r}^{2}h=\pi \left[f\left(x\right){\right]}^{2}\delta x$.
Step 2
Hence, using the fact that for any function g(x) we have we see that the total volume, V, that we are interested in is equal to $\underset{\delta x\to 0}{lim}\sum _{x=a}^{x=b}\pi \left[f\left(x\right){\right]}^{2}\delta x=\pi \underset{\delta x\to 0}{lim}\sum _{x=a}^{x=b}\left[f\left(x\right){\right]}^{2}\delta x$ which is equal to as required. We can apply this very easily to a sphere by considering a circle's Cartesian equation: ${x}^{2}+{y}^{2}={r}^{2}$

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Ethen Blackwell
Answered 2022-07-25 Author has 3 answers
Step 1
Consider a sphere, centred at the origin, with a radius of R. Now think of slicing the sphere into discs of thickness dr. The disc has a surface area of $\pi {r}_{i}^{2}$, where ${r}_{i}$ is the radius of a single disc, so has volume $\pi {r}_{i}^{2}dr$.
Step 2
What is ${r}_{i}$? It is found by realising that ${r}_{i}$ is the horizontal side of a right-angled triangle with R being the hypotenuse and r being the vertical side. Draw it out to convince yourself.
Step 3
To find the volume of the sphere, we have to add up all of the $\pi {r}_{i}^{2}dr$. This is an integral over r, which ranges from -R to R:
$V={\int }_{-R}^{R}\pi \left({R}^{2}-{r}^{2}\right)dr.$

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