Here's the question you clicked on:
mr570
How do I find the integral from -4 to 4 of (3*sqrt(16-x^2)^2 dx ? I think you have to use u-substitution but I don't entirely understand how that works in this case.
well the sqrt gets eliminated with the power 2, so...Its a simple integral
\[y=\sqrt{16-x^2}\] is the upper half of a circle centered at the origin with radius 4 so \[6\int_{-4}^4\sqrt{16-x^2}dx\] is 6 times the area of the upper half of a circle with radius 4
oh sorry i didn't see the square at the end if it is really the square root squared, then it is easy enough, although a rather strange way to write it
it is really \[3\int_{-4}^4\sqrt{16-x^2}^2dx\]?
Assuming it's what sattelite said then this becomes very simple.
|dw:1354850130324:dw| Then this become easy to integrate.
i think the typo from asker