# Thread: Need help with cylindrical shells

1. ## Need help with cylindrical shells

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis.

Problem:
y = 4x^2, 2x + y = 6

I have found out that those two equations intersect at (-1.5, 9) and (1, 4). Since this is bounded by the x-axis, I use the y values (9 and 4) as the limits for the integral.

However this is where a problem comes in. I don't know how to figure out this problem.

I have transformed both equations so that now it is:
x= sqrt(x/4) and x = 3 - (y/2)

Seeing that x = 3 - (y/2) is higher than x= sqrt(x/4) I set up the integral problem as this:

Integral from 4 to 9 (2pi)(y)(3 - (y/2) - sqrt(x/4))

But when I integrate that I get some crazy numbers that does not equal the answer at the back of the book.

The book has (250pi)/3 as the answer.

Please help me with this frustrating problem! Thank you so much in advance.

2. This is why I like to do ALL such problems both ways. Not only does it provide more practice and more challenging problems, but I get a result verification for free.

WASHERS

$\displaystyle \int_{-\frac{3}{2}}^{1}\pi\cdot\left[(6-2x)^{2}-(4x^{2})^{2}\right]\;dx\;=\;\frac{250\cdot\pi}{3}$

Well, that does support the book. You may wish to recognize that when solving for 'x', there are TWO branches that you might need to consider. Somewhere in that square root, we might need the negative value. Keep your eyes open for it.

SHELLS[1]

$\displaystyle \int_{0}^{4}2\cdot\pi\cdot y\cdot (\sqrt{\frac{y}{4}}-(-\sqrt{\frac{y}{4}}))\;dy$

There's the first piece. What do you get for the rest?

SHELLS[2]

$\displaystyle \int_{4}^{9}2\cdot\pi\cdot y\cdot (What(y)-WhatElse(y))\;dy$

3. Originally Posted by TKHunny
This is why I like to do ALL such problems both ways. Not only does it provide more practice and more challenging problems, but I get a result verification for free.

WASHERS

$\displaystyle \int_{-\frac{3}{2}}^{1}\pi\cdot\left[(6-2x)^{2}-(4x^{2})^{2}\right]\;dx\;=\;\frac{250\cdot\pi}{3}$

Well, that does support the book. You may wish to recognize that when solving for 'x', there are TWO branches that you might need to consider. Somewhere in that square root, we might need the negative value. Keep your eyes open for it.

SHELLS[1]

$\displaystyle \int_{0}^{4}2\cdot\pi\cdot y\cdot (\sqrt{\frac{y}{4}}-(-\sqrt{\frac{y}{4}}))\;dy$

There's the first piece. What do you get for the rest?

SHELLS[2]

$\displaystyle \int_{4}^{9}2\cdot\pi\cdot y\cdot (What(y)-WhatElse(y))\;dy$
Hm I don't really get what you are saying in the second part. Would it be 2*pi*y*([(y-6)/2]-(-[(y-6)/2])?

4. No good. Nice try, but you have used the line on both sides. Draw some horizontal lines at y = 6 or y = 7 or y = 8 and see that the height of the segment is the line (as you have it) less the negative branch of the parabola (as it is in the first piece).

[0,4] -- Positive Branch less Negative Branch
[4,9] -- Line less Negative Branch

Give it another go.