Question

can someone help me please
Sketch the region enclosed by x=y^2-6 and x=-3-2y^2. Then find the area of the region.

Answer 1

Well, it's not really necessary to sketch the region, but it does help. Hopefully you can do that part.

These functions are given in terms of y, so we'll have to integrate with respect to y.

If you imagine swapping our axes, such that the x-axis is now vertical, and the y-axis is the horizontal and independent one.

Then, x = y^2 - 6 is the standard parabola shifted downwards 6 units.
x = -2y^2 - 3 is thinner parabola that opens downwards, and is shifted down 3 units from the origin.

Now we need to find their intersection points. So we set them equal to one another.

y^2 - 6 = -2y^2 - 3
3y^2 = 3
y=1 or y=-1.

So, our bounds are from -1 to 1.

Which function is higher? It's x = -2y^2 - 3 since it opens downwards, and is shifted only -3 units, and not -6.

So we integrate [(-2y^2 - 3) - (y^2 - 6)] dy from -1 to 1.

However, this integral is symmetric, so it's just 2∫(-3y^2 + 3) dy from 0 to 1.

This is -y^3 + 3y evaluated at 1 to 0.
=> -(1)^3 + 3(1) - 0 = 2.

We now multiply it by 2 to get 4.

So the area in between those two curves is 4.

Answer 2

I got (-1,1) but i didn't they were symmetric and i think tahts why i was getting -4

Answer 3

Thank YOU!!!

Answer 4

Noticing that they're symmetric isn't at all necessary, it just helps cut down on work, and gives you less opportunities to mess up. You're welcome!