Why there is a difference between the way we calculate surface area and volume using integration?

Suppose we want to find the volume of the curve when rotated about x-axis. We would write something like this: It's like we are considering small discs of radius f(x) and width dx and summing them up as the volume of the disc is pi * f(x) * f(x) * dx

But when we need to calculate surface area, we write something like this: Shouldn't we consider the surface area of a disc which is 2 * pi * f(x) * dx and summing it up rather than 2 * pi * f(x) * small arc length?

#Calculus

Note by Raj Khare
2 years ago

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Comments

Nope.....consider a small cylinder with its center on the X-Axis and the function is being rotated about the X-Axis. Now, we know that it's surface area (curved) is 2πrh\displaystyle 2\pi rh where r\displaystyle r and h\displaystyle h are its radius and height respectively. So, the surface area of our element (small cylinder) is

dS=2πf(x)⋅1+f′(x)2dxdS=2\pi f\left(x\right)\cdot\sqrt{1+f'\left(x\right)^2}dx

because, the height h\displaystyle h can be seen as the distance between two points very close to each other (x1 , y1) and (x2 , y2)\displaystyle \left(x_1\ ,\ y_1\right)\ and\ \left(x_2\ ,\ y_2\right). Computing the distance between them, we get h=(dx)2+(dy)2\displaystyle h=\sqrt{\left(dx\right)^2+\left(dy\right)^2} which implies h=(dx)21+(f′(x))2\displaystyle h=\sqrt{\left(dx\right)^2}\sqrt{1+\left(f'\left(x\right)\right)^2} since dy=f′(x)dx\displaystyle dy=f'\left(x\right)dx and thus, we get from the above formula

S=∫ab2πf(x)1+(f′(x))2dxS=\int_a^b2\pi f\left(x\right)\sqrt{1+\left(f'\left(x\right)\right)^2}dx

where aa and bb are the points on X - Axis between which the area needs to be calculated.

Aaghaz Mahajan - 2 years ago

Then, in the same manner, while calculating the area under the curve we should take the area of small rectangle element = f(x) * root(1 + f'(x)^2)dx?​

Raj Khare - 2 years ago
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