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Find the surface area when this curve is revolved around the x ax
x = 1/3(y^2 + 2)^3/2 [1,2]
I set it up both ways and i get two really complicated integrals.
[tex]2 \pi/3 \int_1^2 \sqrt{\frac{y^2+y^4}{(y^2+2)^{2/3}}} dy [/tex]
Yeah I can't figure out how to do this integral, and I am thinking there must be an easier way. The other way, integrating with x, seemed even more complicated If anyone can give me some hints, it'll be appreciated. Thanks.
x = 1/3(y^2 + 2)^3/2 [1,2]
I set it up both ways and i get two really complicated integrals.
[tex]2 \pi/3 \int_1^2 \sqrt{\frac{y^2+y^4}{(y^2+2)^{2/3}}} dy [/tex]
Yeah I can't figure out how to do this integral, and I am thinking there must be an easier way. The other way, integrating with x, seemed even more complicated If anyone can give me some hints, it'll be appreciated. Thanks.