Question

Find the surface area when the indicated portion of the curve is rotated about the specified axis.
y=cos2x 0<=x<=pi/6 x-axis
ill rewrite in first post.

Answer 1

\[y=\cos2x\] \[0 \le x \le \frac{ \pi }{6 }\]

Answer 2

about the x-axis

Answer 3

\[y \prime = -2\sin2x\]

\[\large \int\limits_{0}^{\frac{ \pi }{6 }}2 \pi \cos2x \sqrt{1+(-2\sin2x)^{2}} dx\]

\[\large \int\limits_{0}^{\frac{ \pi }{6 }}2 \pi \cos2x \sqrt{1+(4\sin^{2}2x)} dx\]

and then from here i am stuck

Answer 4

@amistre64