Find the surface area of y=((x^3)/6) + (1/2x)

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gadav478

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Hello all.
I have problem with find the surface area of a curve.

Problem: Find the surface area of y=((x^3)/6) + (1/2x) revolved on the x-axis from x = 1 to x = 2.

I am not sure if I am doing this correctly.

for y(prime) = (x^2/2) - (1/2x^2)
for (y(prime))^2 = (x4/4) - (1/2) + (1/4x^4)
for [(y(prime))^2] + 1= (x4/4) + (1/2) + (1/4x^4) = 1/4 [((x^2) + (1/x2))]^2

After applying Surface formula I get pi[((x^6/36) + (x^2/12) + (x^2/4) - (1/4x^2))] from 1 to 2

Final Answer: 155pi/48

I wanted to know if my calculations are off after or before integrating. Also, I am not sure if there is a better way to integrate possibly using u-sub.

Thanks for the help. :)
 
Last edited:
gadav478 said:
Hello all.
I have problem with find the surface area of a curve.

Problem: Find the surface area of y=((x^3)/6) + (1/2x) revolved on the x-axis from x = 1 to x = 2.

I am not sure if I am doing this correctly.

for y(prime) = (x^2/2) - (1/2x^2)
Click to expand...
Youj have the derivative of x^3/6 right but the derivative of (1/2)x is wrong.

for (y(prime))^2 = (x4/4) - (1/2) + (1/4x^4)
for [(y(prime))^2] + 1= (x4/4) + (1/2) + (1/4x^4) = 1/4 [((x^2) + (1/x2))]^2

After applying Surface formula I get pi[((x^6/36) + (x^2/12) + (x^2/4) - (1/4x^2))] from 1 to 2

Final Answer: 155pi/48

I wanted to know if my calculations are off after or before integrating. Also, I am not sure if there is a better way to integrate possibly using u-sub.

Thanks for the help. :)
Click to expand...
 
gadav478 said:
Hello all.
I have problem with find the surface area of a curve.

Problem: Find the surface area of y=((x^3)/6) + (1/2x) revolved on the x-axis from x = 1 to x = 2.

I am not sure if I am doing this correctly.

for y(prime) = (x^2/2) - (1/2x^2)
for (y(prime))^2 = (x4/4) - (1/2) + (1/4x^4)
for [(y(prime))^2] + 1= (x4/4) + (1/2) + (1/4x^4) = 1/4 [((x^2) + (1/x2))]^2

After applying Surface formula I get pi[((x^6/36) + (x^2/12) + (x^2/4) - (1/4x^2))] from 1 to 2

Final Answer: 155pi/48

I wanted to know if my calculations are off after or before integrating. Also, I am not sure if there is a better way to integrate possibly using u-sub.

Thanks for the help. :)
Click to expand...

Is your function:

\(\displaystyle y = \displaystyle\frac{x^3}{6} + \frac{1}{2x}\) which should be written as y = x^3/6 + 1/(2*x)

or

\(\displaystyle y = \displaystyle\frac{x^3}{6} + \frac{1}{2}x\) which should be written as y = x^2/2 + 1/2 * x

The way you wrote the function in the post implies the second interpretation.

use PEMDAS to post problems correctly
 
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