Question

In problem 4b-2g, why is there the a^2- factor in the integral? Shouldn't it only be (y^2/a)^2?

Answer 1

I have not looked at the question, but If you mean this\[(\frac{y^2}{a})^2\]
that is equal algebraically to
\[\frac{y^4}{a^2}\]

Answer 2

No, I mean there first \[a ^{2}\] in the equation:

\[\int\limits_{0}^{a} \pi (a^{2} - (y^{2} \div a)^{2})dy\]

Answer 3

I see what your asking now ( I looked at the problem ), the \[a^2\]
is the outside of the washer, only it's basically a washer without the hole in it, like a 'coin', then you subtract the center from it \[(\frac{y^2}{a})^2\]to make it a washer, then you do it multiple times ( because the hole in the middle keeps getting wider, however the coin you are subtracting the center from (to make it a washer), namely \[a^2\]never changes, so you have
\[\pi a^2h\]this is the volume of each solid coin, where h is the height
and then from each solid coin you subtract the inner part
\[\pi(\frac{y^2}{a})^2h\]Now to add all the coins with their subtracted inner pieces together, to get the volume of the total object you have the integral
\[\int\limits_{0}^{a}\pi(a^2-(\frac{y^2}{a})^2)dy\]. I find it helpful to always draw a picture of whats going on rather than rely on symbols for understanding, for this problem I did a little sketch also to help me understand.

Answer 4

Got it, thanks.