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

Question

Isabella Ackerman · Accepted Answer

#color(blue)[S_A=2pi int_-3^3 x/pi^2*sqrt[1+(1/pi^2)^2]*dx==(sqrt(pi^2+1)*9)/pi^3-(sqrt(pi^2+1)*9)/pi^3=0]# If the solid is obtained by rotating the graph of #y=f(x)# from #x=a# to #x=b# around the #"x-axis"# then the surface area #S_A# can be found by the integral #S_A=2pi int_a^b f(x)sqrt{1+[f'(x)]^2}dx# #S_A=2pi int_-3^3 x/pi^2*sqrt[1+(1/pi^2)^2]*dx# #S_A=2 int_-3^3 x/pi*sqrt[1+(1/pi^4)]*dx# #S_A=2 int_-3^3 x/pi*sqrt[(1+pi^2)/pi^4)*dx# #S_A=2/pisqrt[(1+pi^2)/pi^4) int_-3^3 x*dx=[(sqrt(pi^2+1)*x^2)/pi^3]_-3^3# #=(sqrt(pi^2+1)*9)/pi^3-(sqrt(pi^2+1)*9)/pi^3=0#

William Abdalla · Answer

See below

Have erased answer.

Elijah Abram · Answer

undefined 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.