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Sagot :
The volume of the smaller wedge cut from a sphere of radius 4 by two planes that intersect along a diameter at an angle of π/6 is 1/6.
What is volume ?
Volume is a measurement of three-dimensional space that is occupied. It is frequently expressed quantitatively using SI-derived units, as well as several imperial or US-standard units. Volume and the notion of length are connected.
Consider that the tetrahedron is bounded by the coordinate planes and the plane 8 x+y+z=2
Evaluate the triple integral [tex]\iiint_E d V$.[/tex]
The lower boundary of tetrahedron is the plane z=0 and the upper boundary of tetrahedron is the plane z=2-8 x-y.
And the projection is y=2-8 x.
The x-limits are obtained by taking y=0 and z=0 in 8 x+y+z=2.
[tex]$\begin{array}{r}8 x=2 \\x=\frac{2}{8} \\x=\frac{1}{4}\end{array}$[/tex]
Therefore, the region becomes [tex]$R=\left\{(x, y, z) / 0 \leq x \leq \frac{1}{4}, 0 \leq y \leq 2-8 x, 0 \leq z \leq 2-8 x-y\right\}$[/tex].
Therefore, the integral becomes
[tex]$ \iiint_E d V=\int_0^{\frac{1}{4}} \int_0^{2-8 x} \int_0^{2-8 x-y} d z d y d x $[/tex]
[tex]$=\int_0^{\frac{1}{4}} \int_0^{2-8 x}[z]_0^{2-8 x-y} d y d x $[/tex]
[tex]$ =\int_0^{\frac{1}{4}} \int_0^{2-8 x}[2-8 x-y] d y d x $[/tex]
[tex]$ =\int_0^{\frac{1}{4}}\left[2 y-8 x y-\frac{y^2}{2}\right]_0^{2-8 x} d x . $[/tex]
[tex]$ =\int_0^{\frac{1}{4}}\left[2(2-8 x)-8 x(2-8 x)-\frac{(2-8 x)^2}{2}\right] d x $[/tex]
[tex]$ =\int_0^{\frac{1}{4}}\left[32 x^2-16 x+2\right] d x $[/tex]
[tex]$ =\left[32\left(\frac{x^3}{3}\right)-16\left(\frac{x^2}{2}\right)+2 x\right]_0 $[/tex]
[tex]$ =32\left(\frac{\left(\frac{1}{4}\right)^3}{3}\right)-16\left(\frac{\left(\frac{1}{4}\right)^2}{2}\right)+2\left(\frac{1}{4}\right)-0 $[/tex]
[tex]$ =\frac{1}{6} $[/tex]
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