Volume of Rotation: Find Area Rotating Around Line

[tree.lee]

Homework Statement

Find volume when the area rotates around a line.

Homework Equations

x=y^2
x=1-y^2
about x = 3

The Attempt at a Solution

Could someone please check my formula for Area? I have pi[ (3-(y^2))^2 - (2-(y^2))^2 ] which gives me the integral pi[5y-(2/3)y^3]. I think it's wrong though because I plugged it into the wolfram alpha program and it gives me the wrong answer, the answer according to the textbook is (10)(sqrt2)(pi)/3 and the program says according to my integral between the intersection points (-1/sqrt2, 1/sqrt2) the volume would be (14)(sqrt2)(pi)/3. I'm really confused. Any help would be greatly appreciated! Thank you.

 

[Mark44]

Homework Statement

Find volume when the area rotates around a line.

Homework Equations

x=y^2
x=1-y^2
about x = 3

The Attempt at a Solution

Could someone please check my formula for Area? I have pi[ (3-(y^2))^2 - (2-(y^2))^2 ]

You are missing either $\Delta x$ or $\Delta y$, which makes it harder for me to tell whether you are using disks or shells. I'm sure I could figure out what you're doing with some effort, but I shouldn't have to.

which gives me the integral pi[5y-(2/3)y^3]. I think it's wrong though because I plugged it into the wolfram alpha program and it gives me the wrong answer, the answer according to the textbook is (10)(sqrt2)(pi)/3 and the program says according to my integral between the intersection points (-1/sqrt2, 1/sqrt2) the volume would be (14)(sqrt2)(pi)/3. I'm really confused. Any help would be greatly appreciated! Thank you.

 

[tree.lee]

You are missing either $\Delta x$ or $\Delta y$, which makes it harder for me to tell whether you are using disks or shells. I'm sure I could figure out what you're doing with some effort, but I shouldn't have to.

Oh, I think I'm using delta y. Like my widths are on the y axis? Is that what is meant? Sorry I'm not very good with terminology at all, haha.

 

[Brian T]

The radius of your circles run parallel to the x-axis, and the summing of the areas occurs along the y-axis axis (the circles are vertically stacked), which gives us
∫\pi(x(y))^2Δy
In this case, x(y) is not directly explicit. You have to use the given functions and the fact that it revolves around the x=3 axis to determine the real form of x(y)

 

[Mark44]

I get the same number you got -- $\frac{14 \pi \sqrt{2}}{3}$. Double-check the problem information and answer. If everything is as you reported, it seems that the book has a typo.

 

[Mark44]

The radius of your circles run parallel to the x-axis, and the summing of the areas occurs along the y-axis axis (the circles are vertically stacked), which gives us
∫\pi(x(y))^2Δy

The above is incorrect. tree.lee is using "washers", or disks with holes in them.

In this case, x(y) is not directly explicit.

Please read the problem more carefully. Both functions give x explicitly in terms of y.

You have to use the given functions and the fact that it revolves around the x=3 axis to determine the real form of x(y)

 

[Dick]

Homework Statement

Find volume when the area rotates around a line.

Homework Equations

x=y^2
x=1-y^2
about x = 3

The Attempt at a Solution

Could someone please check my formula for Area? I have pi[ (3-(y^2))^2 - (2-(y^2))^2 ] which gives me the integral pi[5y-(2/3)y^3]. I think it's wrong though because I plugged it into the wolfram alpha program and it gives me the wrong answer, the answer according to the textbook is (10)(sqrt2)(pi)/3 and the program says according to my integral between the intersection points (-1/sqrt2, 1/sqrt2) the volume would be (14)(sqrt2)(pi)/3. I'm really confused. Any help would be greatly appreciated! Thank you.

3-(1-y^2) does not give 2-y^2.

 

[tree.lee]

The radius of your circles run parallel to the x-axis, and the summing of the areas occurs along the y-axis axis (the circles are vertically stacked), which gives us
∫\pi(x(y))^2Δy
In this case, x(y) is not directly explicit. You have to use the given functions and the fact that it revolves around the x=3 axis to determine the real form of x(y)

Oh. Wouldn't subtracting the functions from x=3 give the distance from it and thus the radii desired? Then subtracting the inner function from the outer function should give the area I want to integrate, right? That gives me ∫ pi[5y-(2/3)y^3], doesn't it?

I get the same number you got -- $\frac{14 \pi \sqrt{2}}{3}$.

But the textbook says (10)(sqrt2)(pi)/3.

 

[Mark44]

But the textbook says (10)(sqrt2)(pi)/3.

Yes, I know. It might be a typo in the book. That has been known to happen.

What book are you using?

 

[Brian T]

I believe I got the answer. Let me type out my work and someone correct me if I'm wrong

 

[tree.lee]

Oh. Wouldn't subtracting the functions from x=3 give the distance from it and thus the radii desired? Then subtracting the inner function from the outer function should give the area I want to integrate, right? That gives me ∫ pi[5y-(2/3)y^3], doesn't it?But the textbook says (10)(sqrt2)(pi)/3.

OH I

Oh. Wouldn't subtracting the functions from x=3 give the distance from it and thus the radii desired? Then subtracting the inner function from the outer function should give the area I want to integrate, right? That gives me ∫ pi[5y-(2/3)y^3], doesn't it?But the textbook says (10)(sqrt2)(pi)/3.

Sorry I posted replies that I didn't mean to I'm new to this forum and there are a lot of brackets saying quotes and stuff and it confuses me and I'm not sure what I"m inputting when I hit post reply. Sorry. Thanks for the help A LOT! I did double check everything.

 

[Dick]

Oh. Wouldn't subtracting the functions from x=3 give the distance from it and thus the radii desired? Then subtracting the inner function from the outer function should give the area I want to integrate, right? That gives me ∫ pi[5y-(2/3)y^3], doesn't it?But the textbook says (10)(sqrt2)(pi)/3.

No. See my post #7.

 

[Mark44]

No. See my post #7.

Yes, good eye, Dick -- you're right. The inner radius is (3 - (1 - y²)², which is (2 + y²)². Using that gives the textbook's answer.

 

[tree.lee]

Yes, good eye, Dick -- you're right. The inner radius is (3 - (1 - y²)², which is (2 + y²)². Using that gives the textbook's answer.

Oh. Wow, this community is so cool. Thanks a billion. It's seriously appreciated. Hope you're all having a wonderful day. Thanks a lot Dick.