Hi, I have the area [tex]D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1[/tex] That is rotated about the x axis, and i need to calculate the area [tex]\pi \int_0^1 3^2-y^2 = \pi \int_0^1 9-xe^{2x^2}[/tex] [tex]\frac{-9\pi}{4}\cdot (e^{2x^2}-1)\bigg|_0^1[/tex] But this is all wrong, why?
Hi James889! First, are you trying to find the area of a surface of revolution, or the volume of a solid of revolution? Anyway, how did you get 9/4 out of that? Do the two parts separately (you seem to be suffering from a sort of human-fly syndrome ).
Hi Tim, Im trying to find the area of a solid. I just factored out the 9 from the integral, the [tex]1/4[/tex] is from integrating [tex]xe^{2x^2}[/tex]
Yes, but how did they get toegther? Anyway … You mean the surface area? But you're using ∫πy^{2}dx, which is a volume. For the correct formula, see http://en.wikipedia.org/wiki/Surface_of_revolution.
Turned out i had misread the question, they did ask for the volume of the solid [tex] D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1 [/tex] Same as before, baby steps. [tex]\pi\int_0^1 9-(\sqrt{x}e^{x^2})^2[/tex] [tex]9x-\frac{e^{2x^2}}{4}\bigg|_0^1[/tex] [tex]\pi\cdot\frac{36-e^2}{4} - (0-\frac{e}{4})[/tex]
Fine, except the last term should be e^{0}/4, = 1/4. (and use more brackets, to show you have the π in the right place)