1. ## Math Review Help

Question 1: Solved

Find the lateral(side) surface area of the cone generated by revolving the line segment y = x/2, 0<= x <= 4, about the x-axis.

i got (pi*sqrt5)/2 just like the image however the answer is 4pi*sqrt5

now when i looked at the image they multiply my answer with

x^2/2

my question is how did they get that and why did they multiply it into the answer...

Question 2: how do i get

$x = y^4/4 + 1/8y^2$

in terms of y =
and find the derivative of it

2. Originally Posted by Reminisce
Question:

Find the lateral(side) surface area of the cone generated by revolving the line segment y = x/2, 0<= x <= 4, about the x-axis.

i got (pi*sqrt5)/2 just like the image however the answer is 4pi*sqrt5

now when i looked at the image they multiply my answer with

x^2/2

my question is how did they get that and why did they multiply it into the answer...
$\frac{x^2}{2}$ is the antiderivative of $x$

$A = 2\pi \int_0^4 \frac{x}{2} \sqrt{1 + \frac{1}{4}} \, dx$

$A = \frac{\sqrt{5}}{2} \pi \int_0^4 x \, dx$

$A = \frac{\sqrt{5}}{2} \pi \left[\frac{x^2}{2}\right]_0^4$

$A = \frac{\sqrt{5}}{2} \pi \left[\frac{4^2}{2} - 0\right] = 4\sqrt{5} \pi$

3. oh my other question then is, for problems like this one do i always have to do that?

4. Originally Posted by Reminisce
oh my other question then is, for problems like this one do i always have to do that?
if you want to determine a surface area of revolution, you do.

5. Question 2.. how do i get

$x = y^4/4 + 1/8y^2$

in terms of y =

6. Originally Posted by Reminisce
Question 2.. how do i get

$x = y^4/4 + 1/8y^2$

in terms of y =
first of all ...

is it $x = \frac{y^4}{4} + \frac{1}{8y^2}$

or $x = \frac{y^4}{4} + \frac{1}{8} y^2$

?

second ...

why do you need to get y in terms of x? what is the context of this problem?

7. i was looking at the solution manual but it turns out that the manual skip alot the question and didn't fix it... But thanks..