I have to find the Surface Area of the solid that is created by the curve: y=1-x^(2)
rotated around the y-axis.
on the interval 0 less than or equal to x less than or equal to 1

Question

I have to find the Surface Area of the solid that is created by the curve: y=1-x^(2)  
rotated around the y-axis. 
on the interval 0 less than or equal to x less than or equal to 1

anonymous · Answer

so far I have $$2\pi \int\limits_{0}^{1}$$

anonymous · Answer

$$\int\limits_{a}^{b} 2pif(x)*\sqrt(1 + (dy/dx)^2)$$

anonymous · Answer

I brought the 2pi out because it's a constant

anonymous · Answer

actually, let me correct that, the f(x) will just be x because your x value will be the radius of the solid

anonymous · Answer

so you should get: $$2\pi \int\limits_{0}^{1} x \sqrt{1 + 4x^2}dx$$

anonymous · Answer

ahhh, that makes sense

anonymous · Answer

Thanks!

anonymous · Answer

no problem