$$x^2/4+y^2/2=1$$
The curve is rotating around the x-axis and I'm suppose to calculate the area.
My attempt: $$y=\sqrt{2-x^2/2}$$
The integral goes from -2 to 2. I tried to simplify the integral and if I didn't make any mistakes, the expression within the integral is $\sqrt{2-(3/4)x^2}$. But I don't know how to solve it from there.
Sorry for my poor formatting skills.
Your help would be very appreciated!
If you really do have $\sqrt{2-\frac{3x^{2}}{4}}=\sqrt{2}\sqrt{1-\frac{3x^{2}}{8}}$, then I would set $x = \sqrt{\frac{8}{3}}\cos\theta$ so that $dx = -\sqrt{\frac{8}{3}}\sin\theta d\theta$, and then integrate over the correct range in $\theta$.