What is the surface area of the solid created by revolving #f(x)=x^2# for #x in [1,2]# around the x-axis?

Question

Isaiah Allen · Accepted Answer

#2pi[1/8 17^(3/2) -1/16 sqrt 17 -1/64 ln|sqrt 17 +4| -1/16 5^(3/2) +1/32 sqrt 5 +1/64 ln |sqrt 5 +2|]#

Now, integrating by parts #int sec^3 theta=int sec theta sec^2 theta d theta#

=#sectheta int sec^2 theta d theta - int sec theta tan theta int sec^2 theta d theta #

=#sec theta tan theta -int sec theta tan^2 theta d theta#

=#sec theta tan theta - int sec^3 theta d theta +int sec theta d theta#

Now transposing #int sec^3 theta d theta #to the left side and integrating #sec theta d theta#,

#2int sec^3 theta d theta= sec theta tan theta +ln|sec theta +tan theta|#

Thus #int sec^3 theta d theta=1/2 sec theta tan theta +1/2 ln |sec theta + tan theta| #

Like wise, using technique of integration by parts #int sec^5 theta d theta= int sec^3 theta sec^2 theta d theta = sec^3 theta int sec^2 theta d theta - int (3sec^2 theta sec theta tan theta) int (sec^2 theta) d theta#

=#sec^3 theta tan theta-3 int sec^3 theta tan^2 theta d theta#

=#sec^3 theta tan theta -3int (sec^5 theta -sec^3 theta d theta)#

Now transpose #int sec^5 theta d theta# to the right side to get

#int sec ^5 theta d theta = 1/4 sec^3 theta tan theta + 3/4 int sec^3 theta d theta#

Now using the integral of #sec^3 theta# derived earlier,

#int x^2 sqrt (4x^2 +1) dx= 1/8[ int sec^5 theta d theta - int sec^3 theta d theta]=1/8[ 1/4 sec^3 theta tan theta -1/4 int sec^3 theta d theta]#

=#1/32[sec^3 theta tan theta -1/2 sec theta tan theta -1/2 ln|sec theta +tan theta|]#

Now substituting theta by x it would be #tan theta =2x# and #sec theta= sqrt(4x^2 +1)#

#int x^2 sqrt (4x^2 +1) dx = 1/16 x (4x^2+1)^(3/2)-1/32 x sqrt (4x^2+1) -1/64 ln |sqrt (4x^2 +1) +2x| #

The required surface area would thus be

#2 pi[ 1/16 x (4x^2+1)^(3/2)-1/32 x sqrt (4x^2+1) -1/64 ln |sqrt (4x^2 +1) +2x|]_1^2 #

=#2pi[1/8 17^(3/2) -1/16 sqrt 17 -1/64 ln|sqrt 17 +4| -1/16 5^(3/2) +1/32 sqrt 5 +1/64 ln |sqrt 5 +2|]#

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Aurora Alvarado · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = x^2 $ for $ x $ in the interval $[1,2]$ around the x-axis, you can use the formula for the surface area of a solid of revolution:

$$ S = 2\pi \int_{1}^{2} f(x) \sqrt{1 + \left(\frac{df}{dx}\right)^2} \, dx $$

Where $ \frac{df}{dx} $ represents the derivative of $ f(x) $ with respect to $ x $. In this case, $ f(x) = x^2 $, so $ \frac{df}{dx} = 2x $.

Substituting the values into the formula:

$$ S = 2\pi \int_{1}^{2} x^2 \sqrt{1 + (2x)^2} \, dx $$

Now, you can integrate this expression within the given bounds to find the surface area of the solid.

Paisley Johnson · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = x^2 $ for $ x $ in the interval $[1,2]$ around the x-axis, you can use the formula for the surface area of a solid of revolution, which is given by:

$$ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx $$

where $ f(x) $ is the function being rotated, and $ a $ and $ b $ are the limits of integration.

In this case, $ f(x) = x^2 $, $ a = 1 $, and $ b = 2 $. The derivative of $ f(x) $ is $ f'(x) = 2x $. Plugging these into the formula:

$$ S = 2\pi \int_{1}^{2} x^2 \sqrt{1 + (2x)^2} dx $$

You can evaluate this integral to find the surface area.