So, I was taught that if we have to find the component for a very small change in volume say $dV$ then it is equal to the product of total surface of the object say $s$ and the small thickness say $dr$.

For example let us take a sphere of radius $r$, then by this method, $$dV = 4πr^2dr$$

Now, if we use calculus, then, $$V = \frac{4}{3}πr^3$$ $$dV = 4πr^2dr$$

Now both the expressions are equal and this the same case for a cylinder as well, but for a cube of side length say $r$ the expression by the method comes out to be, $$dV = 6r^2dr$$

Now by the calculus method, $$ V = r^3$$ $$ dV = 3r^2dr$$

Now, we have two different expressions for $dV$ however this should not be possible in maths as both the expressions are not equal to each other.

Am I doing something wrong here or am I missing something? And if I ever have to take the component in physics should I always go with the earlier method and not the calculus approach?

So, I was taught that if we have to find the component for a very small change in volume say $dV$ then it is equal to the product of total surface of the object say $s$ and the small thickness say $dr$.

For example let us take a sphere of radius $r$, then by this method, $$dV = 4πr^2dr.$$

Now, if we use calculus, then, $$V = \frac{4}{3}πr^3$$ $$dV = 4πr^2dr$$

Now both the expressions are equal and this the same case for a cylinder as well, but for a cube of side length say $r$ the expression by the method comes out to be, $$dV = 6r^2dr.$$

Now by the calculus method, $$ V = r^3$$ $$ dV = 3r^2dr.$$

Now, we have two different expressions for $dV$ however this should not be possible in maths as both the expressions are not equal to each other.

Am I doing something wrong here or am I missing something? And if I ever have to take the component in physics should I always go with the earlier method and not the calculus approach?