Jon feafe's questions at Yahoo Answers regarding volumes by slicing

[MarkFL]

Here are the questions:

I need calculus math help?


Find the volume V of the described solid S.
The base of S is an elliptical region with boundary curve 16x^2 + 25y^2 = 400. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

Find the volume V of the described solid S.
The base of S is the region enclosed by the parabola
y = 3 - 2x^2
and the x−axis. Cross-sections perpendicular to the y−axis are squares.

I have posted a link there to this thread so the OP can view my work.

edit: This question has since been deleted at Yahoo! Answers.

 

[MarkFL]

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:

$$V=2\left[3y-\frac{1}{2}y^2 \right]_0^3=2\cdot\frac{9}{2}=9$$