Respuesta :
Using polar/cylindrical coordinates, we take
x = r cos(t )
y = r sin(t )
z = z
so that
dV = dx dy dz = r dr dt dz
The solid is given by the set
D = {(x, y, z) : x ² + y ² ≤ 1 and 0 ≤ z ≤ 4 - 2x - y}
or in cylindrical coordinates,
D = {(r, t, z) : 0 ≤ r ≤ 1, 0 ≤ t ≤ 2π, and 0 ≤ z ≤ 4 - 2r cos(t ) - r sin(t )}
Then the volume of the solid is
[tex]\displaystyle \iiint_D \mathrm dV = \iiint_Dr\,\mathrm dr\,\mathrm dt\,\mathrm dz \\\\ = \int_0^{2\pi} \int_0^1 \int_0^{4-2r\cos(t)-r\sin(t)} r\,\mathrm dz\,\mathrm dr\,\mathrm d t \\\\ = \int_0^{2\pi}\int_0^1 (4r-2r^2\cos(t)-r^2\sin(t))\,\mathrm dr\,\mathrm dt \\\\ = \int_0^{2\pi} \left(2 - \frac23 \cos(t) - \frac13 \sin(t)\right) \,\mathrm dt \\\\ = \boxed{4\pi}[/tex]
The volume of the given solid below the plane 2x + y + z = 4 and above the disk x2 + y2 ≤ 1 is 4π.
What is the volume of the solid in polar coordinates?
The volume of the solid in polar coordinates can be calculated by the double integral in polar coordinates.
[tex]V = \iint_R f(x,y) \,dA.\\[/tex]
Similarly, by polar coordinates,
x = r cost
y = r sint
z = z
we get, dV = dx dy dz = r dr dt dz
The solid is given
D = {(x, y, z) : x ² + y ² ≤ 1 and 0 ≤ z ≤ 4 - 2x - y}
or in polar coordinates,
D = {(r, t, z) : 0 ≤ r ≤ 1, 0 ≤ t ≤ 2π, and 0 ≤ z ≤ 4 - 2r cos(t ) - r sin(t )}
The volume of the solid is
[tex]\rm \int \int\int\ dV = \int \int\int\ rdr dt dz\\ \rm = \int\limits^{2\pi}_0\int\limits^1_0 \int\limits^{4-2rcost-rsint}_0 rdr dt dz\\\rm =\int\limits^{2\pi}_0\int\limits^1_0 {4r-2r^{2} cost-r^{2} sint} dr dt\\\rm =\int\limits^{2\pi}_0( 2-2/3cost-1/3sint) dt\\\rm =4\pi[/tex]
Hence, The volume of the given solid below the plane 2x + y + z = 4 and above the disk x2 + y2 ≤ 1 is 4π.
Learn more about the volume of the solid;
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