Please note that the equilateral triangle cross-section is parallel to $YZ$ plane (perpendicular to $x-$axis). The base of the equilateral triangle is $x^2$ as it is between $y = 0$ and $y = x^2$.
So sides of equilateral triangle $ = x^2$ and height $ = \frac{\sqrt3}{2} x^2$
For a given value of $z$, $y$ will vary between $\frac{z}{\sqrt3}$ and $x^2-\frac{z}{\sqrt3}$.
The limits of $y$ come from the fact that at any given height ($z$) in the triangle if we drop perpendicular from two edges to the base, the distance from both vertices of the base to the foot of the perpendicular will be $ = z \cot 60^0= \frac{z}{\sqrt3} \ $.
So the integral to find volume should be
$\displaystyle \int_{0}^{4} \int_{0}^{\frac{\sqrt3}{2} x^2} \int_{\frac{z}{\sqrt3}}^{x^2- \frac{z}{\sqrt3}} dy \ dz \ dx$
If you integrate wrt $dz$ first, you will have to split the integral into two hence the choice of the order.
THe method of using integraion to find find volume that you are using is
- Pick an axis.
- Slice the volume in slices perpendicular to that axis.
- Integrate the volume of the slices with respect to the coordinate along that axis.
(Generally, the axis need not be a coordinate axis, but it is convenient when this happens.)
When the problem has
For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the x-axis and one radius in the $xy$-plane.
You are told that the slice is a quarter circle, so has one quarter the area of a circle, and that (1) all the radii for that quarter circular are perpendicular to the $x$-axis and (2) one radius is in the $xy$-plane. This means that in a slice, one can take the distance perpendicular to the $x$-axis to the line $y = 2x$ as the radius of the circle, one quarter of which area is accumulated into the integral. (In the $xy$-plane, a perpendicular to the $x$-axis is parallel to the $y$-axis. A ray starting on the $x$ axis, in the first quadrant of the $xy$-plane, and proceeding perpendicular to the axis crosses the triangle base of the volume, reaching the end of the base at the line $y = 2x$. (If this is not clear, sketch a graph of the base.)
Suppose $r$ is the radius of a circle. The area of that circle is $\pi r^2$. The area of one-quarter of that circle is $\frac{1}{4} \pi r^2$. The distance from the $x$-axis to the line $y = 2x$ (the radius described by the ray in the previous paragraph) is $2x$, so the radius we are to take for the slice at coordinate $x$ is $r = 2x$. Then the area contribution from the slice at coordinate $x$ is $\frac{1}{4} \pi (2x)^2$. That is the integrand that we integrate from $x = 0$ to $x = 2$. (Again, if it is not clear that this is the range of coordinate that includes all the slices, sketch the base.)
The following diagram of the volume is from a point in the $(+++)$-octant looking back throught the volume at the origin.
Example slices (for $x \in \{1/2, 1, 3/2, 2\}$ are shown as fairly clear quarter circles. The base is the $xy$-plane, at the bottom of the diagram. The dashed gray line is in the $xz$-plane and marks the ends of the radii of the quarter circles in that plane.
We can more clearly see the "slicing" viewing from the $(+-+)$-octant nearly parallel to the positive $y$-axis.
This angle reinforces that we are given the axis, we are given the area formula for the slices, and all we have to find is the relation between the $x$-coordinate and the area of the corresponding slice.
Best Answer
You have to find volume of the area bound between curves $C1$ and $C2$ when rotated around $x = 2$
$C1: y=\frac{\pi}{4} x$ and $C2: y=\arctan x$ $(0 \leq x \leq 1, 0 \leq y \leq \frac{\pi}{4}$)
When you rotate it around $x = 2$, please note the radius of the rotation of both curves at any given $y$.
For $C1$, $r = 2 - \frac{4y}{\pi}$
For $C2$, $r = 2 - \tan y$
So your triple integral to find volume should be $\displaystyle \int_0^{2 \pi} \int_0^{\pi/4} \int_{2 - \frac{4y}{\pi}}^{2 - \tan y} r \ dr \ dy \ d\theta$