algebra-precalculuscalculusdefinite integralsderivativesintegration
So I was given the following prompt:
"What is the volume of the solid generated when the region in the first quadrant bounded by the graph of $y=x^3$, the $y$-axis, and the horizontal line $y=1$ is revolved about the $y$-axis?"
I guess I'm just a bit confused about what the integral set up might look like here. I understand the basics of the Disc and Washer methods, but I'm a bit confused about how that would apply here. Any help would be appreciated!
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
(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 can find intersection of $y=1, y = x^3$ which is $(1,1)$. So the integral to find volume can be set up using washer method as
$0 \leq x \leq y^{1/3} \ $ where $x$ is the radius around $y-$axis
$0 \leq y \leq 1$
$\displaystyle \int_0^1 \int_0^{y^{1/3}} 2 \pi x \ dx \ dy$
Or it can be set up using shell method as
$x^3 \leq y \leq 1$
$0 \leq x \leq 1$
$\displaystyle \int_0^1 \int_{x^3}^1 2\pi x \ dy \ dx$