Each plane perpendicular to the x-axis intersects a certain solid in a circular cross section whose diameter lies in the xy-plane and extends from $x^2 = 4y$ to $y^2 = 4x$. The solid lies between the points of intersection of these curves. Find its volume
My approach
$\mathcal The$ $\mathcal graph$
$y^2=4x \Rightarrow y = 2\sqrt{x} = f(x)$
$x^2=4y \Rightarrow y = {x^2\over 4} = g(x)$
$\pmb The$ $\pmb Solid$
Points of intersection are (0,0) and (4,4) In [0,4] , f(x) > g(x)
So radius at x , $2r = f(x) - g(x)$
$r = \sqrt{x} - {x^2\over 8}$
Volume of disc $dv = \pi r^2dx$
$= \pi [x + {x^4\over 64} - {x^2\sqrt{x}\over 4}]dx$
$\int_{0}^{V}dv = \int_{0}^{4} \pi [x + {x^4\over 64} - {x^2\sqrt{x}\over 4}]dx$
$= \pi [ {x^2\over 2} + {x^5\over 5×64} - {x^{7\over 2}\over 14} ]_{0}^{4}$
$= {72\pi \over 35}$
I don't know whether my approach and steps are correct or not pointing out the error in my solution and suggesting the right one would be of much help
Your integration/thinking process is correct however you made minor calculation errors along the way. For starters when writing g(x), you simplify wrong by dividing by 2 instead of 4. So it should have been $g(x) = x^2/4$. Additionally, when writing the radius in terms of $x$ as you miswrite the $g(x)$ function again as $(x/2)/2$ when in reality it should have been $(x^2/4)/2 = x^2/8$