Volumes of solids of revolution?

[ahat]

Find the volume of the solid of revolution formed when the region between the y-axis and the curves of y =cos^-1(x) and y = sin^-1(x) is rotated about the y-axis.

I don' think I'm approaching it correctly. I did V = ∫ cos²(y) - sin²(y) dy between 0 and but obviously this equals 0. I don't think I'm interpreting the question correctly.

Help appreciated.

[b^3]

Do a quick sketch first, here's one I prepared earlier

Now we are rotating around the axis, which in this case means you need to split the integral into two regions, adding the volumes formed by rotating the red curve from to the intersection of the curves (remember since we have this will be the corresponding coordinate) and then rotated the blue curve from there to . This means you will need to find the point of intersection of the two curves (check your value for that, and then remember to take the value).

Spoiler

Hope that helps

[ahat]

Thanks heaps b^3 I'm still a bit confused as to why the integral needs to be split? Why is it that the
V =  π ∫ [f(y)]² - [g(y)]² dy doesn't hold in this case? Apologies for my ignorance.

[b^3]

That rule would work if the region we wanted was between the two curves, and , but in our case the region is not between those curves, rather it's between either of those curves and the axis. In other words we are rotating the shaded region below around the axis.

Now we have the region bounded by a different curve depending on the value. For it's bounded by the red curve and the axes while for it's bounded by the blue curve and the axis. So we can split this region up into two regions, and rotate them separately, and then add the volumes of those two regions.

[ahat]

Oh my gosh b^3 hahaha. So simple, why did I not get that. Thanks so much, +1 respect,

[lzxnl]

With EVERY SINGLE area or volume question, draw a rough sketch of just exactly what you're finding first. Trust me, it'll paradoxically save you time and marks lost.