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Find the area bounded by $x= at^{2}$ and $y = 2at$ from $t=1$ to $t=2$

I tried to solve this by integrating $\int_{1}^2 y \frac{dx}{dt} dt$

$\int_{1}^2 (4a^{2}t^{2}) dt$

$= (28/3)a^{2}$

What is wrong with my attempt$?$

Answer is given $(56/3)a^{2}$

Mathaddict
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    Well, for one, your integration is wrong: the antiderivative of your integral (taking $a$ constant) is

    $$\int 4at^2 dt = 4a \int t^2 dt = \frac 4 3 at^3 + C$$

    No $a^3$ anywhere. (Also $dx/dt = 2at$ so you'd actually have $a^2$.) Of course the problem lies deeper but I figured it worth noting.

     – PrincessEev Aug 09 '19 at 11:47
  • @Mathsaddict: How is that a bounded region? – quasi Aug 09 '19 at 11:48
  • I think we're supposed to assume it is bounded on the left by $x=a$ and on the right by $x=4a$? – Cameron Williams Aug 09 '19 at 11:50
  • And on the bottom? – quasi Aug 09 '19 at 11:51
  • Ah true - $x$ axis presumably. – Cameron Williams Aug 09 '19 at 11:52
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    That should be stated. – quasi Aug 09 '19 at 11:53
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    I don't disagree on that. – Cameron Williams Aug 09 '19 at 11:53
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    For OP: if you get tripped up on these, just straight up solve $y$ in terms of $x$ and fix your limits of integration. Then see if you can find where your mistake might have come from when working with the implicit $t$ integral. – Cameron Williams Aug 09 '19 at 11:55
  • @CameronWilliams This question is given in a book exactly as I stated here. I googled this problem before asking and there are many other sites I could find for the solution of same question. But I didn't get how did they take the limits for solution – Mathaddict Aug 09 '19 at 12:01
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    @Mathsaddict: Assuming the lower boundary was intended as the $x$-axis, your answer is correct. Note: answer keys can be wrong. – quasi Aug 09 '19 at 12:04
  • Somewhere I saw, that this is a parabola and limits were from $x=a$ to $x=2a$ and from $y= - \sqrt{4ax}$ to $y= \sqrt{4ax}$ – Mathaddict Aug 09 '19 at 12:04
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    It's not a full parabola. Note that if $a > 0$, then $x,y$ are both positive when $1\le t\le2$, so you only get the upper half of the parabola. Presumably, that's where whoever made the answer key got it wrong. – quasi Aug 09 '19 at 12:06
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    Oh geez this is a case of someone taking a square root and using both solutions when both are not necessarily correct in the context of the broader problem.. – Cameron Williams Aug 09 '19 at 12:10
  • @EeveeTrainer $y \cdot (dx/dt) = 2at \cdot 2at = 4a^2t^2$? – Ahmed Hossam Aug 09 '19 at 12:42
  • @AhmedHossam Yes, that's correct? Not sure what you're trying to show/ask me. If you're wondering why my post has $a$ not $a^2$, OP's post had $a$ before an edit, but that's all I can guess – PrincessEev Aug 09 '19 at 12:44
  • That must be it then, an edit. – Ahmed Hossam Aug 09 '19 at 12:45

3 Answers3

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Your answer is just fine, because $$\int_{1}^{2} 4a^2 ~dt = \int_{a}^{4a} 2a\sqrt{\frac{x}{a}}~ dx = 28a^2/3$$ is the area between the parametric curve and the $x(t)$-axis. enter image description here

Ahmed Hossam
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Hint: You can also use an explicit form of your function: $$t=\sqrt{\frac{x}{a}}=t$$ so $$y=2a\sqrt{\frac{x}{a}}$$

Dr. Sonnhard Graubner
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The curve is a parabola y^2 = 4aX area will be twice the area under curve y=sqrt.(4ax) from X= a to X = 4a and x axis answer is 56 a^2 / 3

mathsdiscussion.com
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    $$\int_{x=a}^{x=4a}\sqrt{4ax}~dx=\int_{x=a}^{x=4a}2\sqrt{ax}~dx=\int_{x=a}^{x=4a}2a\sqrt{\frac{x}{a}}~dx=28a^2/3$$ – Ahmed Hossam Aug 09 '19 at 20:53