# Calculate the surface area with integration

Calculate the surface area of the surface obtained when the region enclosed by the given curves is revolved about the $$x$$-axis $$y=2x^2-8$$ $$y=x^2-1$$

This is a model problem for an exam and I really don't know what to do. I don't have any idea what shape will I get when the given curves are revolved around the x-axis. Can somebody help me with this problem?

Can I consider these areas for the calculation

• You should first draw the curves and find where they interesect. [The shape you get before you revolve arund the x-axis will be*] a sort of inverted vee-shaped thing with curved edges, I expect. But you should draw it for yourself. Edit: [*] – Zubin Mukerjee Oct 13 '18 at 16:34
• @J.Dane Nice problem! – Michael Rozenberg Oct 13 '18 at 16:36
• An upside-down symmetrical version of the star trek logo! – Zubin Mukerjee Oct 13 '18 at 16:37
• I drew it but I don't know how should I seperate the integral in order to calculate the surface because I don't think that the shape I get when I revolve the curves can be calculated with one integral – J.Dane Oct 13 '18 at 16:39
• I want the surface area of the surface obtained. Am I saying it right because English isn't my native language? – J.Dane Oct 13 '18 at 16:49

I won't do the whole problem but the $$3$$ surfaces are on the intervals $$[0, 1.7320508],$$
$$[1.7320508, 2.6457513]\ \text{and} \ [2, 2.6457513]$$ whereby symmetry about the y axis means doubling each area to get the total area.