# Revolving Asymmetric Function Around Y-Axis Where It is Bonded From -Ve Value to +Ve Value.

So I am facing this problem where I don't know which method to use to find the volume of a rotating function:

$y= \sec(x) \sqrt{\tan(x)+1}$

from $x=-\fracπ4$ to $x=\fracπ4$ around the y-axis.

As you may know, the left and the right sides of the function don't look the same, meaning that the solid will be irregular. I tried dividing the problem into two parts:

1- Find the volume of the rotated right hand side (from $x=0$ to $x=\fracπ4$) around the y-axis then divide it by 2.

2- Find the volume of the rotated left hand side (from $x=-\fracπ4$ to $x=0$) around the y axis then divide it by 2.

3- Add the two volumes to find the total volume of the solid formed.

But apparently, this method turned out to be wrong as one of my peers solved it differently (in a method I dont understand) and the teacher confirmed it to be right.

• You are counting the intersection of the area you obtained from parts 1 and 2 twice. Commented Oct 11, 2015 at 6:15
• what do you mean the intersection? could you elaborate please? Commented Oct 11, 2015 at 6:20
• The figure obtained from the rotated right hand side will intersect with the figure obtained from the rotated left hand side, hence you cannot find the volume simply by adding them up. Commented Oct 11, 2015 at 6:23
• Then how can I begin to solve it? The function is not symmetric about the y axis so we can't use cylindrical shells or disks method.. Commented Oct 11, 2015 at 8:03

You use the formula $$V=\pi\int_{x=a}^{b}y^2\mathrm{d}x$$ where $V$ is the volume of revolution for rotation about the $x$-axis.
Or $$V=\pi\int_{y=a}^{b}x^2\mathrm{d}y$$ for rotation about the $y$-axis