<p>Not refering to anything specifically, but how do you integrate an circle with cross sections of a square?</p>
<p>Well the side of the square would be 2y so then the area of the square would be (2y)^2 = 4y^2 so you would substitute in for y in terms of x and just integrate that from the left to right.</p>
<p>(2y)^2 = 4y^2</p>
<p>^how did u get that?</p>
<p>(2y)^2 = (2y)(2y) = 4*(y)^2</p>
<p>the shape you are making cross-sections for is a square (a=s^2).
In this case, since the circle is centered at 0, one side of the square would equal 2y.
y would be the equation of the circle in terms of x <a href=“2y”>in other words, f(x)</a>^2 = 4y^2
then, take the integral of 4f(x)^2 on whatever interval the circle is on.</p>