proving the area of a heart

[teal perch]

choufleur

[edgy canopy]

oh man i'm not great with calculus helping

idk how much help people give in forums but id rec using the question channels lol

if you dont get help here that is

[tight rock]

y_s is the centerline, so it's root x

It's just saying that the height is equal to twice the distance to the midpoint

[peak tulip]

In part one of the solution they separate the x^2 to the other side of the equation. Next we take the sqrt of both sides. The left side will now be y - sqrt(x), and by using the fact that the sqrt(x) is y_s (the midpoint between both y_1 and y_2) we can rewrite it as y_1 - y_s. To get the height we simply need to just multiply the y_1 - y_s by 2. Because we multiplied the left side by 2 to transform the left side we need to multiply the left side by 2 as well. We know that we need to integrate over the distance between y_2 and y_1 equation so we setup our bounds to be -sqrt(2) to sqrt(2) but we also know that the equation is even therefore we can change our bounds to 0 to sqrt(2) and simply multiply the result by 2. We also know that y_2 - y_1 is equal to 2(y_2 -y_s) which is equal to 2sqrt(2-x^2). Lastly by pulling out the 2 inside our integral to the outside portion and multiplying by the 2 we already had previously pulled we are left with our final integral 4 times the integral which has the bounds 0 to sqrt(4) over sqrt(2 - x^2) dx

[vernal orbit]

thanks, i get it now :)

[tight rock]

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[vernal orbit]

.solved