Thread: Find the area enclosed by the curve and the y-axis

1. Find the area enclosed by the curve and the y-axis

I have to find the area enclosed by the curve: x = t2-2t, y = t1/2 and the y-axis. I know that this is some kind of intergral, but I don't really know how to approach it because it is on the y-axis.

How would I setup the integral?

Thank you.

2. Re: Find the area enclosed by the curve and the y-axis

Solving for t in terms of x would be ugly, because t shows up twice in that equation. Rather, solve for it in terms of y using the second equation.

if $y = t^{\frac{1}{2}}$

then

$y^2 = t$

$x = y^4 - 2y^2 = y^2(y^2-2)$

This is convenient for us because have an expression for x in terms of y, perfect for finding the area between this and the y axis.

$A = |\int_{0}^{\sqrt{2}} y^4 - 2y^2 dy|$

3. Re: Find the area enclosed by the curve and the y-axis

Question: Why is the limit from 0 to sqrt(2)?

I ended up with area = (4sqrt(2))/5) - (4sqrt(2)/3) -- Is that correct?

4. Re: Find the area enclosed by the curve and the y-axis

Originally Posted by Shadow236
Question: Why is the limit from 0 to sqrt(2)?

I ended up with area = (4sqrt(2))/5) - (4sqrt(2)/3) -- Is that correct?
The limit is from 0 to $\sqrt{2}$ because those are the points at which your graph intersects the y-axis - think about it, it intercepts the y-axis when $x=0 \implies y^2(y^2 -2) = 0 \implies y = 0, \ \sqrt{2}$ $\big($we discard $y = - \sqrt{2}$ because y is defined by $y = \sqrt{t} \implies y > 0 \ \forall t \in \mathbb{R}$ $\big)$

Yes but remember, area is a positive quantity:

$\text{Area} = \left| \int_{0}^{\sqrt{2}} y^2 (y^2 - 2) \ dy \right|$

5. Re: Find the area enclosed by the curve and the y-axis

Originally Posted by FelixFelicis28
$\big( y > 0 \ \forall t \in \mathbb{R}$ $\big)$
Scratch that, meant $y \geq 0 \ \forall \ t \geq 0$

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find the area enclosed by the curve x = t2 âˆ’ 2t, y = t and the y-axis.

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