- #1
stunner5000pt
- 1,461
- 2
Homework Statement
The given curve is rotated about the y-axis. Find the area of the resulting surface.
[tex] y = 3 - x^2 [/tex]
[itex] 0 \leq x \leq 4 [/itex]
Homework Equations
[tex] A_{y} =2\pi\int_a^b x \sqrt{1+\left(\frac{dx}{dy}\right)^2} dy [/tex]
The Attempt at a Solution
now we need to write x in terms of y so we get:
[tex] x = \pm \sqrt{3-y} [/tex]
and if we differentiate the above, we get:
[tex] \frac{dx}{dy} = -\frac{1}{\sqrt{3-y}} [/tex]
Our range also changes
when x = 0, then y = 3
and x = 4, then y = -13
so then our integral is:
[tex] A_{y} =2\pi\int_{-13}^{3} 2\sqrt{3-x} \sqrt{1+ \left(-\frac{1}{\sqrt{3-y}}\right)^2} dy [/tex]
So i multiplied the above by 2 instead of plus or minus
and in addition to that, do we need to consider the dx/dy term? Does that need to be multiplied as well?
Thanks for your help!