Question

Integration problem that was posted/solved earlier, but now trying to solve it using a different method. Finding the volume of an object using the shell method (One moment).

Answer 1

"Find the solid's volume generated by revolving the region bounded by the x-axis, the curve y = 3x^4, and the line x = 1 and x = -1 about the x-axis."

I chose to do this using the shell method and integrating with respect to y in the following way:

Answer 2

The general formula for the shell method is\[V = \int\limits_{a}^{b}2 \pi(r)(h)dx\]Where r is the shell radius and shell height. This time, the thickness variable is instead dy, and basically any term that can be will be expressed in y. I found my new integration bounds my plugging in for x = 1, which should have the same y-value as x = -1, and plugged in for 0 for the lower bound, which should by default be zero:
\[b = 3(1)^{4} = 3\]\[a = 3(0)^{4} = 0\]

Answer 3

From there, I found the circumference of the circle of the cylinder that you would depict for the shell method, in terms of y:\[C = 2 \pi y\] I then found how to express the cylinder width(length) in terms of y, which should be \[(2- \sqrt[4]{\frac{ y }{ 3 }})\]Because \[y = 3x ^{4}, x = \sqrt[4]{\frac{ y }{ 3 }}\]

Answer 4

Let me just cut to the chase by now:
\[V = \int\limits_{a}^{b}2 \pi (r)(h)dx = 2 \pi \int\limits_{0}^{3}(y)(2 - \sqrt[4]{\frac{ y }{ 3 }})dy \]\[V = 2 \pi \int\limits_{0}^{3}(2y - \sqrt[4]{\frac{ y ^{2} }{ 3 }})\]And then upon integrating it and evaluating it, it just becomes a total mess, but the answer is just 2pi. What am I doing wrong here?

Answer 5

Oh, and I was using this as a guide: https://www.khanacademy.org/math/calculus/solid_revolution_topic/shell-method/v/shell-method-for-rotating-around-horizontal-line

Answer 6

why do you have \(2-\sqrt[4]{\frac{y}{3}}\) (i'm focusing on the 2)

Answer 7

\[2\int\limits_{0}^{3}2\pi y\left(1-\sqrt[4]{\frac{y}{3}}\right)dy=2\pi\]

Answer 8

Sorry, my computer just couldn't even handle this rad calculus being done apparently and decided to overheat all of a sudden. No, the 2 inside is to represent the bounds of integration in terms of y. The total range (butchering the terms for now, but it doesn't really matter) on the x-axis in this problem equal 2 (-1 through 1), and the way you could write...ugh, it's hard to describe this, all you have to do is watch about four minutes into the video and you'll see what I mean.

Answer 9

@Zarkon

Answer 10

you cant do that...2 is not -1 through 1

Answer 11

Can you describe how to do it, then? lol.

Answer 12

It's the same in terms of being the absolute value of a distance, which is what it looked like I could do from the video; I don't really have any other examples to go off, none where it goes into the negative portion of the x-axis.