The base is the semicircle $$y=\sqrt{16−x^2},$$ where -4 $\le$ $x$ $\le$ 4. The cross-sections perpendicular to the $x$-axis are squares. $$\\$$ So far this is what I have:
$\int (Area)\,dx$
$\implies$ $\int \frac{π}{2} ($$r^2$$)\ dx$.
$r$ = $\frac{\sqrt{16−x^2}}{2}$
$\implies$ $ \frac {π}{8}$ $\int 16- $$x^2$$\ dx$. $$\\$$ I'm confused with what I have to do with the information given about the squares.
On
Each slice of the solid is a square.
The side length of the square is equal to the height of the function at that point.
So each square has area $(\sqrt{16-x^2})^2=16-x^2$.
You just need to integrate $\displaystyle \int_{-4}^4 (16-x^2)\,dx=\boxed{\frac{256}{3}}$.
Note, the base is the region bounded by $y=\sqrt{16-x^2}$ AND $y=0$.
Therefore, the side of the square at each point would be the distance between the two curves.
On
You should be considering the slice which is square in shape. Imagine a vertical line drawn at some point $x$ that meets the semi-circle. This will be the side of your square slice. Note the length of this line is simply the $y-$value of the point where the line meets the semicircle.
Now compute the area of this square slice as $y^2$. Let $dx$ be the thickness of this slice.
The volume of this square slice is $y^2 \, dx$.
So the total volume is $$\int_{-4}^{4} y^2 \, dx.$$
You must use the area of a square.
Remember that the area of a square is $s^2$.
For this problem, your side length is $y=\sqrt{16−x^2}$, which means your area for the square is $y=\sqrt{16−x^2}^2.$ $$\\$$
$$\int_{-4}^{4}{16 +x^2 } \,dx$$ $$\\$$ Can you continue from there?