Find the surface of the generated volume

[sun1234]

Homework Statement

The region bounded by the graphs of the curves y=4√x , y=0, x=5, x=12 is revolved about the x-axis.
a) Find the religion lateral surface area of the generated volume.
b) Find the total surface area of the generated volume, including both ends.

Homework Equations

The Attempt at a Solution

Part a:
https://lh6.googleusercontent.com/kZIjTDFGufXqICfEMW2J0CpXFuQ8-OfRJlg91DQJRWZlCpsiCzfND45AYeYCfasTzAUca7ZFa9k=w1256-h843

Part b:

https://lh3.googleusercontent.com/jguorhs3qoad_UbPduGEyBG79Y9WVLG0vPzC4YV6Y1sTNV-gbxczDdBbPKr-G5jIH3wO0g=w1256-h843

Thank you so much for taking time.

 

[Mark44]

Homework Statement

The region bounded by the graphs of the curves y=4√x , y=0, x=5, x=12 is revolved about the x-axis.
a) Find the religion lateral surface area of the generated volume.
b) Find the total surface area of the generated volume, including both ends.

Homework Equations

The Attempt at a Solution

Part a:
https://lh6.googleusercontent.com/kZIjTDFGufXqICfEMW2J0CpXFuQ8-OfRJlg91DQJRWZlCpsiCzfND45AYeYCfasTzAUca7ZFa9k=w1256-h843

You have a mistake in the integral on the next to the last line of your first page. The radius is y, not x. Also, you have a typo where you wrote $y = \frac 2 {\sqrt{x}}$. That should have been y', the derivative.

Part b:

https://lh3.googleusercontent.com/jguorhs3qoad_UbPduGEyBG79Y9WVLG0vPzC4YV6Y1sTNV-gbxczDdBbPKr-G5jIH3wO0g=w1256-h843

Thank you so much for taking time.

 

[fourier jr]

The formula for arc length is $2\pi\int{y\sqrt{1 + (y\prime)^{2}}\ dx}$ which using that, I got that the arc length is $8\pi\int{\sqrt{x+4}\ dx}$. The radius r changes with x!

 

[sun1234]

You have a mistake in the integral on the next to the last line of your first page. The radius is y, not x. Also, you have a typo where you wrote $y = \frac 2 {\sqrt{x}}$. That should have been y', the derivative.

I'm sorry. The question is correct. 2/√x is the derivative of 4√x.

 

[Mark44]

I'm sorry. The question is correct. 2/√x is the derivative of 4√x.

Yes, I understand that. My comment was that you wrote $y = \frac 2 {\sqrt{x}}$ when you should have written $y ' = \frac 2 {\sqrt{x}}$

More importantly, your integral is wrong. It should be
$2\pi \int_5^{12} y\sqrt{\frac{x + 4}{x}}dx$

 

[sun1234]

Why is the radius y? I thought the radius is x or (x-5), I'm not not so sure though.

 

[fourier jr]

For a given x the cross-sections of the solid you're interested in are circles whose centre is at x & whose radii are $4\sqrt x$.

 

[Mark44]

Why is the radius y? I thought the radius is x or (x-5), I'm not not so sure though.

The curve is being rotated around the x-axis, so the radius is y.

 

[sun1234]

The curve is being rotated around the x-axis, so the radius is y.

Oh yes, you're right. So I have to change the whole equation then.

For a given x the cross-sections of the solid you're interested in are circles whose centre is at x & whose radii are $4\sqrt x$.

I really appreciate your help.