When ever you use calculus you need to start with the correct mathematical relationships. You further need to ensure that you integrate over the correct limits and over the correct variables.
You can see integration as the summation of a lot of intesimal small items to give the final answer.
All correctly setup integrals will result in the same answer.
Before one start or rush into solving a problem, think what would be the easiest way to address the problem.
I thought approaching it as a triangle would be quite easy to solve the problem. So this solution will use a triangle as the basic mathematical relationship.
We all know the area of a triangle is:
$$Area = \frac{1}{2}Base * Height$$
I will thus setup the problem as follows. (Solution 1)
Define a triangle with a height of S (defined as the length from point to side of base) and a base of an infinite small length of the base and call it dbase.
As we know that radians is defined in such a way that the circumference of a circle will yield exactly $$2.\pi$$ for a circle of unity (r=1). This relationship helps us to define dbase then as a plain ratio of an infinitely small angle multiplied by the radius. In this case the radius of the cone. Thus we can say:
$$dbase = R. d angle$$
The area of my triangle would then become:
$$darea = \frac{1}{2} S. R. dangle$$
We will integrate between the limits 0 and 2pi to include the total circumference. In degrees this would have been 0 to 360, however the relationship we use is defined for radians.
(Note that I couldn't get the upper limit as $$2\pi$$ and thus integrate over half the circumference and multiply the result by 2. I am not familiar with the syntax. The answer however is the same. In some instances we use this as a trick to reduce the effort in calculating the result of integrals etc. )
$$Area = 2\int_0^ \pi \frac{1}{2} S. R. dangle$$
$$Area=2.(\frac{1}{2} S. R . \pi)-2.(\frac{1}{2} S. R . (0))$$
$$Area =\pi S R$$
And the result is exactly as the rest of the world believe the answer should be.
Important is to be always mathematically correct.
In answers above you made basic mathematical errors and end up with wrong answer.
If you follow maths you can't go wrong. If you went wrong you made a mistake, search your assumptions and relationships and try gain.
Try the volume yourself. I did this on my phone and it is somewhat difficult.
Solution 2 - integrate from 0 to H, where H is height of the cone and R the radius.
h =height, dheight is very small increase in height and r is the radius at h.
Angle is the angle between the vertical and side length S.
The mathematical relationship for the area is the circumference multiple by the small increase in the length of S. Note that the length is not dh but do because of the angle. Be very careful here.
$$darea = 2\pi . r. ds$$
To setup the integral correct we need to express r and ds in terms of h to ensure consistency. The relationship is as follows:
$$tan(angle) = r /h $$
$$r = h. tan(angle) $$
And
$$cos(angle) = dh/ds $$
$$ds = dh. Cos(angle) $$
darea becomes then
$$darea =2\pi.tan(angle). cos(angle) .h.dh$$
We just need to integrate this as:
$$Area = 2\pi. tan(angle). cos(angle) \int_0^ H h. dh$$
$$Area = 2\pi. tan(angle).cos(angle) (( \frac{1}{2} H^2)-(\frac{1}{2} (0)) $$
$$Area = \pi. tan(angle).cos(angle) H^2$$
We only need to get rid of the tan and cos and then we should have the same answer. From the above we can substitute back to get:
$$Area = \pi. S. R$$
It worked again, no magic.
Solution 3 - integrate from 0 to S. Same variables as in solution 2.
Area definition same. This time we need to express r in terms of s.
$$darea = 2\pi . r. ds$$
$$sin(angle) = r /s $$
$$r = s. sin(angle) $$
Replace r in darea, then integrate
$$Area = 2\pi. sin(angle) \int_0^ S s. ds$$
$$Area = 2\pi. sin(angle) (( \frac{1}{2} S^2)-(\frac{1}{2} (0)) $$
$$Area = \pi.sin(angle) S^2$$
Get rid of the sin term and can you believe, once again the same result as the rest of the world.
$$Area = \pi. S. R$$
I love maths.