more help (volume - integration) (1 Viewer)

[sasquatch]

Im stuck on two integration questions, but i guess if see how one of them is answered i might understand how to do the other:

Find the volume of the solid when the region between the curves y = x^2 and y = 2 - x^2 in the x-y plane is rotated about the y axis.

I sketched and found the two curves are between y = 0 and y = 2. But for some reason when i try to find the volume i get 0?!?

V=π∫²₀(2 - y - y)dy
=π∫²₀(2 - 2y)dy
=π[2y - y²]²₀
=π[4-4]
= 0?

I dont get what im doing wrong... Thanks for any help.

Another question im stuck on:

Find the equation of the tangent to the parabola y = 2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area between the parabola, the tangent line and the x axis is revolved about the x-axis.

Equation of the tangent: y = 4x -2, Point of intersection (1/2,0)

I get 2pi/15 for the answer but the book gives 2pi/3.

Thanks for any help (again).

 

[KeypadSDM]

V = pi₀∫²R²(y)dy

Firstly, find the intersection points of the graphs:

x² = 2 - x²
x = +1
I.e y = 1

Thus, we have to separate the integrals.

For 0 <= y <= 1 we have:

R(y) = x = Sqrt[y]

For 1 <= y <= 2 we have:

R(y) = x = Sqrt(2 - y)

Thus we need the following integral:

V = pi₀∫²R²(y)dy
= pi₀∫¹ydy + pi₁∫²(2 - y)dy
=pi[y²/2]₀¹ + pi[2y - y²/2]₁²
=pi(1/2 - 0 + 2(2) - 4/2 - 2(1) + 1/2)
=pi units cubed.

EDIT: I stuffed up royal!

 

[sasquatch]

I dont get what you did.. would you mind explaining?

 

[KeypadSDM]

Ok, that's just dumb.

How is the radius of the function equal to 2 - y - y?

Seriously, the total volume is twice the volume from 0 to 1 [clearly, just draw a diagram, it's 2 parabolas.

Then the integral becomes x squared dy, which is just ydy.

Hence 2 * pi/2 = pi

 

[sasquatch]

Um well........i didnt say the radius was 2 - y - y... My book explained that the volume of an area between two curves rotated about an axis, is the volume of one curve rotated about one axis, minus the volume of the other curve rotated about the same axis. Thats what i followed. It didnt say anything about the radius..or yeah.. Sorry for bugging you or something but i guess if you get fed up and stuff you dont have to help or anything.

 

[insert-username]

Um well........i didnt say the radius was 2 - y - y... My book explained that the volume of an area between two curves rotated about an axis, is the volume of one curve rotated about one axis, minus the volume of the other curve rotated about the same axis. Thats what i followed. It didnt say anything about the radius..or yeah.. Sorry for bugging you or something but i guess if you get fed up and stuff you dont have to help or anything.

When you're rotating about the x-axis, you need to use functions of x and dx. When you rotate around the y-axis, you use functions of y and dy. So for your first question, you need to change the subject of the functions to x, since you're rotating about the y axis.

y = x²

x = √y

y = 2 - x²

x = √(2 - y)

Give it another try and see how you go. Check also where the two curves intersect when written in this form.


I_F

 

[sasquatch]

yeah i did that...

V=€ˆ20(2 - y - y)dy
=€ˆ20(2 - 2y)dy
=€[2y - y2]20
=€[4-4]
= 0?

If you see 2 - y - y

y = x2

x = √y

y = 2 - x2

x = √(2 - y)

x² = y and x² = 2-y

But what im confused about is i did the volume of one curve rotated about the axis, minus the volume of the other curve rotated about the axis. My book explains that the volume is found like that, but finding the volumes of each seperately, they end up being the same for each curve, hence why i get 0 for my answer. Is that method therefore wrong, or... well Keypad mentioned something about a radius so yeah. Also i looked in the cambridge book and theres not really any explination either.

 

[insert-username]

But what im confused about is i did the volume of one curve rotated about the axis, minus the volume of the other curve rotated about the axis. My book explains that the volume is found like that, but finding the volumes of each seperately, they end up being the same for each curve, hence why i get 0 for my answer. Is that method therefore wrong, or... well Keypad mentioned something about a radius so yeah. Also i looked in the cambridge book and theres not really any explination either.

Sorry, I completely misread that. But I now know what you did wrong.

EDIT: See my post below


I_F

 

[sasquatch]

Hey um i think you misread wrong again..

V=€ˆ20(2 - y - y)dy
=€ˆ20(2 - 2y)dy
=€[2y - y²]20
=€[4-4]
= 0?

Thats what i did

integrating 2 - 2y you get 2y - 2y²/2 which gives 2y - y² which is what i did. Anyway the answer isnt 2pi its just pi.

 

[insert-username]

Dammit! I'm having a very bad night tonight...

It's your limits that are the real problem (yes, the right one this time I know my problem is blindness). You need to use the limits y = 0 to y = 1 (the y-values, since it's a rotation about the y-axis), which will give you pi as an answer. Sorry for all the confusion.


I_F

 

[sasquatch]

Hehe yeah i kow the points of intersection are x = -1 and x = 1 BUT the y values for the points of intersection are y = 1, y = 1. And for this question its a rotation about the y-axis, so arent you supposed to use the y values?

 

[insert-username]

The right solution is: since this is a rotation about the y-axis, you need to first find the integral bound by the curves and the y-axis in one direction, up or down, and then rotate that integral. From the graph, the limits are y=0 and y=1 (the negative part is irrelevant since we're spinning the integral around).

So the integral is:

Pi₀∫¹(2 -2y) dy

= Pi[2y- y²]¹₀

= Pi(1)

= Pi units³

Finally!


I_F

 

[sasquatch]

Um the graphs that you drew were wrong. You drew x^2 = y - a sideways parabola.

x^2 = y is exactly the same as y = x^2 so it is a standard parabola. If you use the x-values x = 1 and x = -1 you end up getting either 2pi or 4 pi..i cant remember. Im really confused now REALLY confused...

 

[insert-username]

Um the graphs that you drew were wrong. You drew x^2 = y - a sideways parabola.

x^2 = y is exactly the same as y = x^2 so it is a standard parabola. If you use the x-values x = 1 and x = -1 you end up getting either 2pi or 4 pi..i cant remember. Im really confused now REALLY confused...

Sorry, it's fixed now. But it's also right now. Your limits are y = 0 and y = 1 (this is a rotation, so the part below the y-axis doesn't matter). Not -1 and 1.


I_F

 

[zeek]

Alright this is my try at the question...
I hope this works/is correct...

 

[sasquatch]

hey hehe thanks for your attempt. The answer in the back of the book is just pi units³, and also it looks like your method is incorrect. The method i followed didnt work for this question, so i cant advise you on what to use...CUZ I DONT KNOW MYSELF!!

 

[insert-username]

I have just had my epiphany. An integral of 0 is correct - there is a positive side and a negative side. However, this is a volume of revolution. If you visualise spinning the integral around, it will spin through the lower part of the integral. You only need to find the integral for the "top" half, and then when you spin that about the y-axis, it passes through the bottom half (not coloured in). That's why you seem to get an integral of 0 - you have a negative and a positive, but you only need one half of it to work it out. When you work out just one half (0 to 1) following my working above, you get the right answer - pi units³. It took us damn long enough.

Your method was right, sasquatch, you just had to halve it. When you find a solid of revolution, you have to be careful with bits under the axis.

I still get the nagging feeling that something's wrong... but I'm too tired to look for it, the answer's right and the question was sneaky.


I_F

 

[sasquatch]

Hehe wow, thanks for all your help. Glad i didnt get you fed up or anything. So just to confirm, when you rotate about either axis, you always use the x limits?

 

[insert-username]

Hehe wow, thanks for all your help. Glad i didnt get you fed up or anything. So just to confirm, when you rotate about either axis, you always use the x limits?

I'm glad I didn't get you fed up. I made that many mistakes... but I'm always happy to help out - it's good for me as well.

When you rotate about the y-axis, you use the y-limits. When you rotate about the x-axis, you use the x-limits, just as in standard integration. I got that backwards too in a post back there, so I'll fix that now. Y-axis: y-limits. X-axis - x-limits.


I_F

 

[sasquatch]

Hmm if you use the y-limits, why would it need to be halved? Like if you see my attached picture when you rotate i see how you get both sides of the y-axis with half a rotation. But the part that is halved is not based on the y-limits it is the x-limits that are halved. The actual x-limits are x = -1, x = 1, but we are only taking it from x = 0 to x = 1. The limits you used were y = 0 and y = 1, meaning only the bottom half of the area is rotated (as seen also in the attachment). So now im still confused...