A volume by slicing without an x-value

Question

I am quite confused with what I am doing wrong in trying to solve for the problem of finding the volume of a solid that has no $x$ values.

The problem offers region bound by $y=x^4, y=1,$ and $y=6$. The problem asks to find the volume of a solid obtained by rotating it about $x = 6$.

I've made the $R(x) = 6$ and $r(x) = 6-x^4$, which feels right but then I don't know if I mess up the bounds by setting it as 6 and 1. I found something about 1 and -1, but it kept giving me 0 as the final answer.

I've been stuck on this problem for a few hours, and have no idea on how to progress, thank you for any and all help that can be provided!

Editing: I squared both radii to get $R(x) = 36$, and $r(x) = 36-12x^4+x^8$ Then I got $12x^4 -x^8$ subtracting the 36. Integrating into $12x^5/5 - x^8/8$ which I tried solving for 6 to 1 and 1 to -1, but neither seemed to be the answer.

Desmos Graph

Answer 1

See the sketch. As you are rotating about $x = 6$, I recommend using washer method to set up the integral to find volume. The curve $y = x^4$ has two branches - one to the right of y-axis and one to the left.

$ \displaystyle y = x^4 \implies x = \pm y^{1/4}$

$x = y^{1/4}$ is the branch to the right of y-axis and $x = - y^{1/4}$ is the branch to the left of y-axis.

The distance of a point on the left branch from $x = 6~$ is $~ f(y) = 6 + y^{1/4}~$ and of a point on the right branch is $~g(y) = 6 - y^{1/4}~$.

So the integral should be,

$V = \displaystyle \pi \int_1^6 \left[(f(y))^2 - (g(y))^2\right] ~ dy$

Use $~a^2 - b^2 = (a-b) \cdot (a + b)$ for simplifying the integrand.

Answer 2

I would do that as two parts. First y= x^4 is equivalent to $x= y^{1/4}$ except that we have to take x positive and negatve, $x= -y^{1/4}$ and $x= y^{1/4}$.

First rotate $x= -y^{1/4}$, that forms the outside of the 'cup', around x= 6. For each y we are rotating a line from x= 6 to $x= -y^{1/4}$. That has length $y^{1/4}+ 6$ and sweeps out a circle of area $\pi(y^{1/4}+ 6)^2$. Taking the thickness to be dy each disk has volum $\pi(y^{1/4}+ 6)^2 dy$ and the total volume $\pi\int_1^6 (y^{1/4}+ 6)^2 dy= \pi \int_1^6 y^{1/2}+ 12y^{1/4}+ 36 dy$.

Now do the other, that forms the inside of the 'cup', $x= y^{1/4}$. Rotating that about the line x= 6, for each y we are rotating line from x= 6 to $x= y^{1/4}$. That has a length of $y^{1/4}- 6$ and sweeps out a circle of radius $y^{1/4}- 6$ so area $\pi(y^{1/4}- 6)^2$. Taking the thickness to be dy each disk has volume $\pi(y^{1/4}- 6)^2dy$ and the total volume is $\pi\int_1^6 (y^{1/4}- 6)^2 dy= \pi]int_1^6 y^{1/2}- 12y^{1/2}+ 36 dy$.

The get the volume of the 'cup', subtract the "inside" from the "outside": $\pi\int_1^6 (y^{1/4}+ 6)^2 dy= \pi \int_1^6 y^{1/2}+ 12y^{1/4}+ 36 dy- \pi\int_1^6 y^{1/2}- 12y^{1/4}+ 36 dy= \pi\int_1^6 24y^{1/4}dy= \left[\frac{96}{5}y^{5/4}\right]_1^6$.