The region below the curve y = e^x + e^(-x) and the lines x = 0, x = ln4 is rotated 2π radians about the x-axis. Find the volume of the resulting solid.
We can use the formula for a Volume of Revolution: V =π ∫ (e^x + e^(-x))^2 dx, with limits x = 0, x = ln4.Expanding the brackets: (e^x + e^(-x))^2 = e^2x + 2 + e^(-2x).So: V = π ∫ (e^2x + 2 + e^(-2x)) dx = π [ (1/2)e^2x - (1/2)e^(-2x) + 2x ], evaluated with limits x = 0, x = ln4.Substituting in the limits we have:V = π( [(1/2)e^2ln4 - (1/2)e^(-2ln4) + 2ln4] - [(1/2) - (1/2) + 0] ) V = π [ (1/2)(4^2 - 4^(-2)) + 2ln4 ]Evaluating: V = π((255/32) + 2ln4).
RS
Related Maths A Level answers
All answers ▸What is the derrivative (dy/dx) of the equation 2 = cos 4x - cos 2y in terms of x and y?
Integrate 1/(5-2x) for 3≤x≤4
The graph with equation y= x^3 - 6x^2 + 11x - 6 intersects the x axis at 1, find the other 2 points at which the graph intersects the x axis
Find the intergal of 2x^5 -1/(4x^3) -5 giving each term in its simplest form.