The base of a solid is a region in the $1st$ quadrant bounded by the x-axis, the y-axis, and the line $x+2y=8$. If the cross-sections of the solid perpendicular to the x-axis are semicircles, what is the volume of the solid?
Area of a semicircle: $\pi r^2\over 2$
Diameter = $8-x\over 2$.
What's my error?
$\endgroup$1 Answer
$\begingroup$The general formula for finding the volume of a solid via slicing is
A = $\displaystyle \int_{a}^{b} A(x) dx$
where A(x) is the cross-sectional area of the solid. We know that the cross-sections are semicircles perpendicular to the x-axis. To find the area of the typical semicircle, we must first find the typical radius.
For each semicircle, the diameter extends from the x-axis to the line x + 2y = 8. Thus to find the typical diameter we must subtract: $\frac{8-x}{2} - 0 = \frac{8-x}{2} = diameter$. To find the radius we divide by 2:
r = $\frac{8-x}{4}$.
Since the area of a semicircle is $\frac{\pi r^2}{2}$, to get the area of our typical cross-section we just plug in:
A(x) = $\frac{\pi}{2}(\frac{8-x}{4})^2$
Now before we integrate we must find the limits of integration, which are in this case the intersection of the line with the x and y axes. So we have a = 0 and b = 8.
Thus we integrate to get volume:
V = $\displaystyle \int_{0}^{8} \frac{\pi}{2}(\frac{8-x}{4})^2 dx$ = $\frac{16\pi}{3}$
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