Volume of solid, varying angle

Hey, had another I could use a hand with. My guess is that I'm reading the task wrong.

**Problem:**

"Compute the volume of a solid that has height h if its horizontal cross section at any height y above its base is a circular sector having radius a and an angle t = 2*pi*(1-(y/h))."

**Meandering attempts at solution:**

The way I interpret it, we have a rather strange solid, which pinches like the middle of an hour glass at y = h/4 and y = 3h/4, and has the full circular area perpendicular to the height at

y = 0, y = h/2 and y = h.

Well, as far as I am aware, you can generally find the volume of any solid whose cross sectional area A(y) is known as a function of the height y by taking the integral of

A(y)dy, in this case between the limits y = 0 and y = h.

The way I want to read it, the cross sectional area should be

A(y) = pi*a^{2}*cos(t)

I then try to simply solve the aforementioned integral, substituting u = t, but then I seem to run into a wall and I'm not sure if it's just me bungling the integral calculation or if I've misunderstood the wording of the task. According to the key the correct answer should be pi*a^{2}*h/2 cubic units.

Re: Volume of solid, varying angle

The sector is $\displaystyle \frac{t}{2\pi}$ of a whole circle, so its area is $\displaystyle \frac{t}{2\pi}\pi a^2=\frac{1}{2}ta^2$. So you substitute the equation for t and integrate from 0 to h dy. The answer the key gives is correct.

- Hollywood

Re: Volume of solid, varying angle

Well, sure, when you put it like that it's trivial. Weird how you can sometimes stare yourself blind at stuff like this. Anyhows, thanks a lot for the assist!