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    So the question is to work out the volume of the solid formed for the parametric equation x=t^2 + 1 and y=t

    - Background info i guess? When I worked out the area, I used the formula ∫t dx/dt dt which was ∫t 2t dt which simplifies to ∫2t^2 dt. The limits I used were t=2 and t= -2 based off the previous part of the question (which was correct) and i got the correct answer of 32/3 which was also correct

    - Back onto the volume - when i did the volume's formula, i got π∫(t)^2 dx/dt dt which is π∫t^2 2t dt which simplifies to π∫2t^3 dt which when you integrate is π[(2/4)t^4] which is simplified to π[(1/2)t^4].

    TO THE QUESTION I HAVE This is now an even function so I can't use the same limits of t=2 and t= -2 and i have to use t=2 and t=0 and double my answer. My answer is 16π, but the actual answer is 8π. Why do I not have to double my answer for this parametric type? Is it a rule for parametrics? Or am I missing something?

    Thanks a lot!
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    (Original post by Killyouraorta)
    So the question is to work out the volume of the solid formed for the parametric equation x=t^2 + 1 and y=t

    - Background info i guess? When I worked out the area, I used the formula ∫t dx/dt dt which was ∫t 2t dt which simplifies to ∫2t^2 dt. The limits I used were t=2 and t= -2 based off the previous part of the question (which was correct) and i got the correct answer of 32/3 which was also correct

    - Back onto the volume - when i did the volume's formula, i got π∫(t)^2 dx/dt dt which is π∫t^2 2t dt which simplifies to π∫2t^3 dt which when you integrate is π[(2/4)t^4] which is simplified to π[(1/2)t^4].

    TO THE QUESTION I HAVE This is now an even function so I can't use the same limits of t=2 and t= -2 and i have to use t=2 and t=0 and double my answer. My answer is 16π, but the actual answer is 8π. Why do I not have to double my answer for this parametric type? Is it a rule for parametrics? Or am I missing something?

    Thanks a lot!
    Mind showing us a picture of the question? Whilst it is the case that the area spans -2 \leq t \leq 2, it is quite likely the case that they only want the volume for the portion of the curve that spans 0 \leq t \leq 2 but I can't confirm that until I see the exact question in full.
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    (Original post by Zacken)
    Mind showing us a picture of the question? Whilst it is the case that the area spans -2 \leq x \leq 2, it is quite likely the case that they only want the volume for the portion of the curve that spans 0 \leq x \leq 2 but I can't confirm that until I see the exact question in full.
    Yea, okay, this might take a second, i have to use the cd to get the question up
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    (Original post by Killyouraorta)
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    your curve roughly looks like that, right?

    using  \displaystyle V = \pi \int_0^2 2t^3\: \text{d}t should give you your answer

    Edit: sorry ill let zacken take this one
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    (Original post by DylanJ42)
    Edit: sorry ill let zacken take this one
    Nah, you take this one, I gotta rush off; I think she knows the equation you've given, but she wants to know why you don't include the bit from t=-2.
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    Name:  image.jpg
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Size:  503.1 KB This is what it looks like

    yeah i basically want to know why you don't double this answer because thats what you usually do with even functions?
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    (Original post by Killyouraorta)
    This is what it looks like
    Can you see your shaded area? That's the only portion of the curve that you're rotating, that portion is defined over 0 \leq t \leq 2. If the bottom bit was shaded too (corresponding to -2 \leq t \leq 0, then you would double your answer. Does that kinda make sense? I'll see if I can come back in a bit and explain better.

    If the bottom bit was shaded then you'd need to integrate \int_{-2}^2 f(t) \, \mathrm{d}t = 2\times \int_{0}^2 f(t) \, \mathrm{d}t - but it's not, so you don't need to.
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    (Original post by Killyouraorta)
    Name:  image.jpg
Views: 110
Size:  503.1 KB This is what it looks like

    yeah i basically want to know why you don't double this answer because thats what you usually do with even functions?
    the shaded bit is only the half above the x-axis, so you just integrate between t=0 and t=2.
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    (Original post by Zacken)
    Can you see your shaded area? That's the only portion of the curve that you're rotating, that portion is defined over 0 \leq t \leq 2. If the bottom bit was shaded too (corresponding to -2 \leq t \leq 0, then you would double your answer. Does that kinda make sense? I'll see if I can come back in a bit and explain better.

    If the bottom bit was shaded then you'd need to integrate \int_{-2}^2 f(t) \, \mathrm{d}t = 2\times \int_{0}^2 f(t) \, \mathrm{d}t - but it's not, so you don't need to.
    (Original post by HapaxOromenon)
    the shaded bit is only the half above the x-axis, so you just integrate between t=0 and t=2.
    Ahhh right okay, thank you! I just assumed that it would be the same as what I did for area. I'm actually so bad at parametric integration omg.. Thank you!
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    (Original post by Killyouraorta)
    Name:  image.jpg
Views: 110
Size:  503.1 KB This is what it looks like
    okay so look at the diagram I posted (or the one in your question, it doesnt matter), do you see how if you use the limits t=0 to t=2 then rotating that part of the curve only 360 degrees about the x axis will give you the volume? (here you are completely ignoring the part of the curve under the x axis)

    see this https://youtu.be/3oAjcLD34kc?t=25s

    if you were you include the part under the curve in a real life experiment can you see how it wouldn't generate any extra volume, as it would just overlap the shape generated by the t=0 to t=2 part?

    general rule i guess, for these integration type questions with rotating around axis etc avoid crossing picking limits which involve crossing axis like the plague :laugh: limits of t<0 can become a mess if not handled properly.

    you could if you wished rotate the area from t=0 to t=2 around 180 degrees and then rotate the area from t=-2 to t=0 180 degrees and add them together (however watch for the minus sign which appears, hence why you do them separate so you can manually deal with that)
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    (Original post by Killyouraorta)
    Ahhh right okay, thank you! I just assumed that it would be the same as what I did for area. I'm actually so bad at parametric integration omg.. Thank you!
    Name:  fff.jpg
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    I want to show you this quickly in response to the point i made about integrating with limits under the x axis and dealing with the zero manually, because you'll probably need to do it at some stage during C3 or C4. If you were asked to find the area enclosed by the curve sinx and the x axis from 0 to 2pi.

    you might firstly try integrating in one big go;  \displaystyle Area = \int_0^{2\pi} \sin x \: \text{d}x and you'll be surprised when the answer is;  \displaystyle Area = 0 , this is because of the whole over/under the x axis thing and how areas under the x axis are negative, so the areas effectively "cancel each other out".

    then you might try  \displaystyle Area = \int_0^{\pi} \sin x \: \text{d}x + \int_{\pi}^{2\pi} \sin x \: \text{d}x , you've considered above and below separately right? so it all should work out this time, however once again you get  \displaystyle Area = 0 . You may have worked out the areas separately but you once again forgot about the minus sign.

    So what you do is find the area from 0 to pi,  \displaystyle Area_1 = \int_0^{\pi} \sin x \: \text{d}x = 2 and you work out the area from pi to 2pi,  \displaystyle Area_2 = \int_{\pi}^{2\pi} \sin x \: \text{d}x = -2

    you then realise the -2 is only because the area is under the axis and you say;  \displaystyle area = 2 + 2 = 4

    alternatively you could work out the area between 0 and pi and multiply it by 2, both work however the first two methods are incorrect as you forgot about the minus

    similar idea with the volume of revolution etc, always stick to one side of the axis if you can and think of rotating the shape in your head beforehand, so you know whether you need to multiply by 2 or not etc etc
 
 
 
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