Question

Problem Set 11: 4G-5
The answer for 4G-5 gives, what I believe, is too small of an answer (Ans: 13pi/3).
The problem asks to find the surface area of a parabola y=x^2, 0<=x<=4 revolved around the y-axis. When x is 4, y = 16. The answer given is 13pi/3 ~ 13.6. However, if I take a gross under-estimate of the area of a triangle with base 4, and height 16, the answer is 32. Which is already much bigger than the given answer. I then notice that they are taking the integral with respect to y, and only going between 0 and 2. Would we need to take the integral between 0-16? Thanks!

Answer 1

Also, in the solution, then solving for dx/dy the solution guide first sets x=√y

And then it state dx/dy =−1/(2√y). However, this is NOT correct right?! The negative shouldn't be there :? I'm quite confused now D:

Answer 2

Yes, they missed the boat on this one.
They are using the wrong limits for y, which should be 0 to 16.
and as you noted
\[ d \left( x = y^\frac{1}{2} \right) \rightarrow dx = \frac{1}{2} y^{-\frac{1}{2} } = \frac{1}{2 \sqrt{y}}\]

(because we square dx, this latter mistake does not affect the answer)

Answer 3

As an exercise, here is the problem using x rather than y.

First, we note that the differential surface area dA is
\( 2 \pi r\ ds \) where ds the the differential arc length along the curve.
the radius r will be x
and ds is given by
\[ ds = \sqrt{ dx^2 + dy^2 } \]
using y= x^2 we get
\[ dy= 2x \ dx \\ dy^2 = 4x^2 \ dx^2 \]
and
\[ ds = (4x^2 + 1)^\frac{1}{2} \ dx\]
The integral becomes
\[2 \pi \int_0^4 (4x^2 + 1)^\frac{1}{2} \ x\ dx\]
if we use a change of variables: u = 4x^2 +1, du = 8x dx
limits u= 4*0+1 to 64+1 or u= 1 to 65, we have
\[2 \pi \cdot \frac{1}{8} \int_1^{65} u^\frac{1}{2} \ du \\
=\frac{\pi}{4}\int_1^{65} u^\frac{1}{2} \ du \\
=\frac{\pi}{4}\cdot \frac{2}{3} u^\frac{3}{2} \bigg|_1^{65} \\
\frac{\pi}{6}\left(65^\frac{3}{2}-1 \right) \approx 273.87
\]

Answer 4

Phi thank you so much for the help and even posting steps on finding the real answer! That answer actually matches the answer I originally got :)
HUGE help!