# What is the surface area produced by rotating #f(x)=x/pi^2, x in [-3,3]# around the x-axis?

By signing up, you agree to our Terms of Service and Privacy Policy

See below

Have erased answer.

By signing up, you agree to our Terms of Service and Privacy Policy

To find the surface area produced by rotating the function ( f(x) = \frac{x}{\pi^2} ) around the x-axis over the interval ([-3, 3]), you can use the formula for surface area of revolution:

[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} , dx ]

Where ( f'(x) ) denotes the derivative of ( f(x) ).

First, calculate ( f'(x) ) by taking the derivative of ( f(x) ), then plug ( f(x) ) and ( f'(x) ) into the surface area formula, and integrate over the given interval ([-3, 3]).

[ f'(x) = \frac{1}{\pi^2} ]

Now plug in the values:

[ A = 2\pi \int_{-3}^{3} \frac{x}{\pi^2} \sqrt{1 + \left( \frac{1}{\pi^2} \right)^2} , dx ]

Solve the integral, then calculate the surface area.

By signing up, you agree to our Terms of Service and Privacy Policy

- What is exponential growth in environmental science?
- How do you find the particular solution to #ysqrt(1-x^2)y'-xsqrt(1-y^2)=0# that satisfies y(0)=1?
- What is the surface area of the solid created by revolving #f(x)=x^2# for #x in [1,2]# around the x-axis?
- How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #y=x^3/6+1/(2x)# on the interval #1/2<=x<=1# ?
- What is arc length parametrization?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7