This is the questions I have trouble with :
Set up (do not evaluate) the integral that gives the surface area of the surface generated by rotating the curve y=tanhx on the interval (0, 1) around the x-axis.
Anyone who can help? Not really sure how to even begin!
Thanks
Surface area will be y dx dTheta
from x 0 to 1, and theta from 0 to 2PI
Area= Int (Int) tanhx dx dtheta
Area=2Pi(int from 0 to 1) y*sqrt(1+(y')^2)dx
Thanks so much! :)
To find the surface area generated by rotating a curve around the x-axis, we need to use the formula for surface area of revolution. The formula is given as:
\[S = \int{2\pi y \sqrt{1+ \left(\frac{dy}{dx}\right)^2}}dx\]
Here, y represents the function that defines the curve, dy/dx represents the derivative of the function with respect to x, and dx represents an infinitesimal element of arc length along the x-axis.
In this problem, the curve is given by y = tanh(x) and we need to find the surface area generated by rotating this curve around the x-axis on the interval (0, 1).
To proceed, we need to find the derivative dy/dx. The derivative of tanh(x) is obtained as follows:
\[\frac{d}{dx}(\tanh(x)) = \frac{d}{dx}\left(\frac{\sinh(x)}{\cosh(x)}\right)\]
Using the quotient rule, we can differentiate this expression as:
\[\frac{d}{dx}\left(\frac{\sinh(x)}{\cosh(x)}\right) = \frac{\cosh(x)\cosh(x) - \sinh(x)\sinh(x)}{\cosh^2(x)}\]
Simplifying this expression, we get:
\[\frac{d}{dx}(\tanh(x)) = \frac{1}{\cosh^2(x)}\]
Now, we have all the components required to set up the integral for the surface area. Plugging in the values into the formula, we get:
\[S = \int_0^1{2\pi \tanh(x) \sqrt{1+ \left(\frac{1}{\cosh^2(x)}\right)^2}}dx\]
Note that we plug in the curve y = tanh(x) and the derivative dy/dx = 1/cosh^2(x) into the formula.
Although we have set up the integral, it is now your task to evaluate it using appropriate integration techniques.