Integration 2005 6c)ii) (1 Viewer)

[dolbinau]

This is on the 2005 HSC paper.

I don't understand how when you find the volume you don't divide it by two (because only half is shaded). Is it because it's rotated? If I was finding just the area would it be half?

 

[Azreil]

Whether you rotate the entire area, or the half that is shaded, you will still come up with the same volume.

Put it this way: If you had a semi circle, and you used that to cut a sphere of play doh, you'd get the same volume to if you used a circle. You do a complete rotation, which is why you wouldn't come up with a hemisphere.

If you were finding the area of the shaded part, you'd go from y=12->y=(intercept) under y=12-2x^2 + y=(intercept)->y=o under y=x^2

 

[dolbinau]

Regarding Volume - Cool. I get it.

Regarding Area, so between those points would also take into consideration the shaded area is bound by the Y axis?

When you go from y= 0->4 (intercept) of say x^2, why doesn't area of the 2nd quadrant get calculated? (I think this is just the rule but still an explanation would be nice)

 

[Azreil]

I think what you're asking is that why it doesn't take into account the area where y is negative.

I believe it's because an integral is the area bounded by two curves (hear me out). In this case, it is the are bounded by y=x^2 and y=0. You know how there is a rule:
A=∫f(x) dx - ∫g(x)dx
=∫f(x)-g(x) dx

Replace the g(x) with 0 (as it is in this example) and you have
A=∫f(x)

Which is bounded by f(x) and 0.

Does that make sense?