Re: jon feafe's questions at Yahoo! Questions regarding volumes by slicing
Hello jon feafe,
1.) We are given the boundary of the base:
$$16x^2+25y^2=400$$
To express this curve in standard form, we may divide through by $400$ to obtain:
$$\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$$
We can see now that our limits of integration will be from $-5$ to $5$.
The volume of an arbitrary slice is:
$$dV=\frac{1}{2}bh\,dx$$
Since the slice has faces which are isosceles right triangles, we know $b=h$, so we have:
$$dV=\frac{1}{2}b^2\,dx$$
If we let $h$ be the hypotenuse, by Pythagoras we may write:
$$b^2+b^2=h^2$$
$$b^2=\frac{1}{2}h^2$$
Thus, we have:
$$dV=\frac{1}{4}h^2\,dx$$
Now, we see that we must have:
$$h=2y=\frac{8}{5}\sqrt{25-x^2}$$
Hence:
$$dV=\frac{16}{25}\left(25-x^2 \right)\,dx$$
Summing the slices, we may write:
$$V=\frac{16}{25}\int_{-5}^5 25-x^2\,dx$$
Using the even-function rule, this becomes:
$$V=\frac{32}{25}\int_{0}^5 25-x^2\,dx$$
Applying the FTOC, we obtain:
$$V=\frac{32}{25}\left[25x-\frac{1}{3}x^3 \right]_0^5=\frac{32}{5^2}\cdot\frac{2\cdot5^3}{3}=\frac{320}{3}$$
2.) The volume of and arbistrary square slice of side length $s$ is:
$$dV=s^2\,dy$$
where:
$$s^2=(2x)^2=4x^2=6-2y=2(3-y)$$
Hence:
$$dV=2(3-y)\,dy$$
Summing the slices, we have:
$$V=2\int_0^3 3-y\,dy$$
Applying the FTOC, we obtain:
$$V=2\left[3y-\frac{1}{2}y^2 \right]_0^3=2\cdot\frac{9}{2}=9$$
