Questions on Washers and Shells Methods

[bbl]

I stumbled upon this problem in my review and I got stumped.

The given curves/lines are: y = lnx, y = 2, y = -2x + 2. The points of intersection are (e^2, 2), (0, 2), and (1, 0).

The questions go like, "Set up an integral using washers if the solid is rotated about the y-axis," and "Set up an integral using shells if the solid is rotated about y = 3."

So, my first approach is to express the equations in terms of x but the y = 2 line confused me since I don't know how to express it as x. So, how do I go about setting up the integrals? I would know how to do it if the line was x = 2 but it was y = 2.

 

[Dr.Peterson]

I stumbled upon this problem in my review and I got stumped.

The given curves/lines are: y = lnx, y = 2, y = -2x + 2. The points of intersection are (e^2, 2), (0, 2), and (1, 0).

The questions go like, "Set up an integral using washers if the solid is rotated about the y-axis," and "Set up an integral using shells if the solid is rotated about y = 3."

So, my first approach is to express the equations in terms of x but the y = 2 line confused me since I don't know how to express it as x. So, how do I go about setting up the integrals? I would know how to do it if the line was x = 2 but it was y = 2.

For both problems, the element of area should be a horizontal strip, so everything should be expressed in terms of y, not x. So y=2 will become a limit of integration.

On the other hand, if you were working in terms of x, then y=2 is just a constant function of x.

We can help more if you show us your work, including a sketch of the region, and as much as you can of setting up the integral. (And take one problem at a time.)

 

[bbl]

For both problems, the element of area should be a horizontal strip, so everything should be expressed in terms of y, not x. So y=2 will become a limit of integration.

On the other hand, if you were working in terms of x, then y=2 is just a constant function of x.

We can help more if you show us your work, including a sketch of the region, and as much as you can of setting up the integral. (And take one problem at a time.)

So far, this is what I have and I know I'm missing something:

Washers: V = \pi \int_{0}^{2} ((e^y)2 - (2x-2)^2) dy

Shells: V = 2\pi \int_{0}^{2} ((3-y)(e^y+2x-2)) dy

Here is the graph that I got from Desmos since my hand-drawn graph is not very good.

 

[Dr.Peterson]

So far, this is what I have and I know I'm missing something:

Washers: \(V = \pi \int_{0}^{2} ((e^y)^2 - (2x-2)^2) dy\)

Shells: \(V = 2\pi \int_{0}^{2} ((3-y)(e^y+2x-2)) dy\)

Here is the graph that I got from Desmos since my hand-drawn graph is not very good.
View attachment 32437

I've added delimiters and a missing "^" so that the Latex above works.

In both integrals, you neglected to rewrite the equation of the line as a function of y. Otherwise, you are almost there.

 

[bbl]

In both integrals, you neglected to rewrite the equation of the line as a function of y. Otherwise, you are almost there.

Can I ask for an elaboration on this part? Sorry, I just got confused.

 

[Dr.Peterson]

Can I ask for an elaboration on this part? Sorry, I just got confused.

The inner radius in the washer, and the left end of the shell, is the x-coordinate of a point on the line, for a specified value of y. You used -2x+2, which is the y-coordinate of a point with a specified value of x. Solve the equation y=-2x+2 for x!

It should be obvious that your integrals are wrong, because both have a mix of x and y in the integrands.

 

[bbl]

The inner radius in the washer, and the left end of the shell, is the x-coordinate of a point on the line, for a specified value of y. You used -2x+2, which is the y-coordinate of a point with a specified value of x. Solve the equation y=-2x+2 for x!

OH I didn't notice that. I actually solved it on paper but somehow, I still put -2x+2. My bad. Thank you for pointing it out!

Washers: [math]V = \pi \int_{0}^{2} ((e^y)^2 - (-\frac{y}{2}+1)^2) dy[/math]Shells: [math]V = 2\pi \int_{0}^{2} ((3-y)(e^y+\frac{y}{2}-1) dy[/math]

 

[Dr.Peterson]

OH I didn't notice that. I actually solved it on paper but somehow, I still put -2x+2. My bad. Thank you for pointing it out!

Washers: [math]V = \pi \int_{0}^{2} ((e^y)^2 - (-\frac{y}{2}+1)^2) dy[/math]Shells: [math]V = 2\pi \int_{0}^{2} ((3-y)(e^y+\frac{y}{2}-1) dy[/math]

Now you have it.