Answer:
The volume of the solid is π/40 cubic units.
Step-by-step explanation:
Please refer to the graph below.
Recall that the area of a semi-circle is given by:
[tex]\displaystyle A=\frac{1}{2}\pi r^2[/tex]
The volume of the solid will be the integral from x = 0 to x = 1 of area A. Since the diameter is given by y, then the radius is y/2. Hence, the volume of the solid is:
[tex]\displaystyle V=\int_0^1\frac{1}{2}\pi \left(\frac{y}{2}\right)^2\, dx[/tex]
Substitute:
[tex]\displaystyle V=\frac{1}{2}\pi\int_0^1\left(\frac{x^2}{2}\right)^2\, dx[/tex]
Simplify:
[tex]\displaystyle V=\frac{1}{2}\pi \int_0^1\frac{x^4}{4}\, dx[/tex]
Integrate:
[tex]\displaystyle V=\frac{1}{2}\pi \left[\frac{x^5}{20}\Big|_0^1\right][/tex]
Evaluate:
[tex]\displaystyle V=\frac{\pi}{40}\left((1)^5-\left(0\right)^5\right)=\frac{\pi}{40}\text{ units}^3[/tex]
The volume of the solid is π/40 cubic units.
Volume of a solid is the measure of the 3 dimensional space it occupies. The volume of the considered solid is obtained as [tex]\dfrac{\pi}{40} \: \rm unit^3[/tex]
For that, we can try to find infinitesimally small 3-d region's volume, and then integrate that region over the dimensions available to get the total volume of the specified region.
We can also use the fact that continuous curves are almost linear and non-changing in infinitely zoomed region.
The given solid has base bounded by x-axis, [tex]y=x^2[/tex] and 0≤x≤1
Its three dimensional region is along the z axis, for each x, there is a semicircle perpendicular with radius being 'y'.
If we take [tex]dx[/tex]x-axis, then the curve [tex]y=x^2[/tex]cylinder(split from height because of semicircle)) with diameter y, and height [tex]dx[/tex]volume is : [tex]V_{dx} = \dfrac{1}{2} \times \pi (\dfrac{y}{2})^2 \times dx = \dfrac{\pi (x^2)^2}{8}dx = \dfrac{\pi x^4}{8} dx[/tex]
Integrating this for 0≤x≤1, we will get the volume of the three dimensional region needed as:
[tex]V = \int_0^4V_{dx} = \int_0^1 \dfrac{\pi x^4}{8} dx = \dfrac{\pi}{8} [\dfrac{x^5}{5}]^1_0 = \dfrac{\pi 1^4}{40} = \dfrac{\pi}{40}[/tex] (in cubic units).
Thus, the volume of the considered solid is obtained as [tex]\dfrac{\pi}{40} \: \rm unit^3[/tex]
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