# Thread: Surface area of revolution question

1. ## Surface area of revolution question

The arc of the parabola $\displaystyle y^2 = 4x$ joining the points $\displaystyle (0, 0)$ and $\displaystyle (3, 2\sqrt{3})$ is rotated through $\displaystyle 2\pi$ radians about the x-axis. Find the area of the curved surface generated giving the answer in the form $\displaystyle \frac{k\pi}{3}$ where k is an integer.

My attempt at answering it is written down here:

I'm not sure if it's right because if it wants an answer in that form, then it would suggest that the integer k isn't divisible by 3 - otherwise it could be simplified. Through that logic, I believe I have made some kind of mistake. Could you please take a look and suggest any area where I have slipped up?

Thanks for your help and time if you can answer

2. $\displaystyle y = 2\sqrt{x}$

$\displaystyle y' = \frac{1}{\sqrt{x}}$

$\displaystyle (y')^2 = \frac{1}{x}$

$\displaystyle 2\pi \int_0^3 2\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}} \, dx$

$\displaystyle 4\pi \int_0^3 \sqrt{x+1} \, dx$

$\displaystyle 4\pi \left[\frac{2}{3}(x+1)^{\frac{3}{2}}\right]_0^3$

$\displaystyle \frac{8\pi}{3}(8 - 1)$

$\displaystyle \frac{56\pi}{3}$

3. Originally Posted by db5vry
The arc of the parabola $\displaystyle y^2 = 4x$ joining the points $\displaystyle (0, 0)$ and $\displaystyle (3, 2\sqrt{3})$ is rotated through $\displaystyle 2\pi$ radians about the x-axis. Find the area of the curved surface generated giving the answer in the form $\displaystyle \frac{k\pi}{3}$ where k is an integer.

My attempt at answering it is written down here:

I'm not sure if it's right because if it wants an answer in that form, then it would suggest that the integer k isn't divisible by 3 - otherwise it could be simplified. Through that logic, I believe I have made some kind of mistake. Could you please take a look and suggest any area where I have slipped up?

Thanks for your help and time if you can answer
$\displaystyle y^2=4x \rightarrow x=\frac{y^2}{4}$
Let $\displaystyle f(y)=\frac{y^2}{4}$.
Then, the surface area = $\displaystyle 2\pi \int_0^{2\sqrt{3}} y \, \sqrt{ 1 + [ \, f'(y) \, ]^2 } \, \, dy$.
I dont know what is the point of that $\displaystyle \frac{k\pi}{3}$ !

4. Originally Posted by skeeter
$\displaystyle y = 2\sqrt{x}$

$\displaystyle y' = \frac{1}{\sqrt{x}}$

$\displaystyle (y')^2 = \frac{1}{x}$

$\displaystyle 2\pi \int_0^3 2\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}} \, dx$

$\displaystyle 4\pi \int_0^3 \sqrt{x+1} \, dx$

$\displaystyle 4\pi \left[\frac{2}{3}(x+1)^{\frac{3}{2}}\right]_0^3$

$\displaystyle \frac{8\pi}{3}(8 - 1)$

$\displaystyle \frac{56\pi}{3}$
Thanks for your solution! I can see what you've done for the most part, only the part which is in the fourth and fifth lines I can't seem to understand. I see you differentiated it after reducing $\displaystyle y^2$ to $\displaystyle y$ but differentiating it implicitly gives the same thing once simplified - I see what has happened by there.

I cannot see how you have got it into that particular form. Please could you show me what you have done there more closely?

I can see you took the 2 out as a constant to get $\displaystyle 4\pi$ but can't see how it simplifies to $\displaystyle \sqrt{x+1}$.

5. Originally Posted by db5vry
Thanks for your solution! I can see what you've done for the most part, only the part which is in the fourth and fifth lines I can't seem to understand. I see you differentiated it after reducing $\displaystyle y^2$ to $\displaystyle y$ but differentiating it implicitly gives the same thing once simplified - I see what has happened by there.

I cannot see how you have got it into that particular form. Please could you show me what you have done there more closely?

I can see you took the 2 out as a constant to get $\displaystyle 4\pi$ but can't see how it simplifies to $\displaystyle \sqrt{x+1}$.
note ... $\displaystyle \sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$

$\displaystyle \sqrt{x} \cdot \sqrt{1 + \frac{1}{x}} = \sqrt{x + 1}$