Question

How do you find the volume of the solid in the first octant, which is bounded by the coordinate planes, the cylinder ${x^2} + {y^2} = 9$ , and the plane $x + z = 9$ ?

Answer 1

To solve this question, firstly we will apply triple integral for finding the value. And then we will integrate the volume parts by parts separately with the help of the given Cartesian plane and the equation of surface.

Complete step by step solution:
The coordinates of the planes $x + z = 9$ and the surface is ${x^2} + {y^2} = 9$ .
Now, we can use a triple integral to represent the volume as follows:
$v = \int \int {\int _R}dV$
where,
$R = \\{ (x,y,z)|x,y,z > 0;\,{x^2} + {y^2} \leqslant 9;\,z < 9 - x\\}$
And, so we can set up a double integral as follows:
$v = \int _a^b\int _c^df(z)dx\,dy \\\ = \int _a^b\int _c^d(9 - x)dx\,dy\, \\\$
We now determine the limits of integration by a cross section in the $xy - plane$ which is a quarter circle of radius 3 centred at the $O$ , and so we have:
$0 \leqslant x \leqslant \sqrt {9 - {y^2}}$ and $0 \leqslant y \leqslant 3$ .
So, the integral of the volume is-
$v = \int _0^3\int _0^{\sqrt {9 - {y^2}} }(9 - x)dx\,dy$
With the nested integral we evaluate from the inside out, so let’s deal with the inner integral:
\int _0^{\sqrt {9 - {y^2}} }(9 - x)dx \\\ = [9x - \dfrac{1}{2}{x^2}]_0^{\sqrt {9 - {y^2}} } \\\ = 9(\sqrt {9 - {y^2}} ) - \dfrac{1}{2}{(\sqrt {9 - {y^2}} )^2} \\\ = 9\sqrt {9 - {y^2}} - \dfrac{1}{2}(9 - {y^2}) \\\
Now, our double integral becomes:
$v = \int _0^3\\{ 9\sqrt {9 - {y^2}} - \dfrac{1}{2}(9 - {y^2})\\} dy$
And, for this integral we can split into the two parts-
${I_1} = \int _0^39\sqrt {9 - {y^2}} dy$
And,
${I_2} = \int _0^3 - \dfrac{1}{2}(9 - {y^2})dy$
We can just evaluate the second integral to get:
${I_2} = - \dfrac{1}{2}{[9y - \dfrac{1}{3}{y^3}]^3} \\\ \,\,\,\,\,\, = ( - \dfrac{1}{2})\\{ (9)(3) - \dfrac{1}{3}(27) - 0\\} \\\ \,\,\,\,\,\, = - 9 \\\$
And for the first integral we use the substitution $y = 3\sin u$ , which gives:
${I_1} = 9\int _0^3\sqrt {9 - {y^2}} dy \\\ \,\,\,\,\, = 9[y\dfrac{{\sqrt {9 - {y^2}} }}{2} + \dfrac{9}{2}\arcsin (\dfrac{y}{3})]_0^3 \\\ \,\,\,\,\, = 9\\{ (0 + \dfrac{9}{2}\dfrac{\pi }{2}) - (0 + 0)\\} \\\ \,\,\,\,\, = \dfrac{{81\pi }}{4} \\\$
Now, we also observe that the above integral $\int _0^3\sqrt {9 - {y^2}} dy$ represents the area of a quarter circle of radius 3, which therefore has area, $A = \dfrac{1}{4}\pi ({3^2}) = \dfrac{{9\pi }}{4}$ which again gives ${I_2} = 9A = \dfrac{{81\pi }}{4}$ .
So, after combining both the parts, the volume as:
$v = \dfrac{{81\pi }}{4} - 9 \\\ \,\,\, = 54.617251.... \\\$
Hence, the volume of the solid in the first octant is $54.617251...$

Note:
An octant in solid geometry is one of the eight divisions of an Euclidean three-dimensional facilitating framework characterized by the indications of the directions. It is like the two-dimensional quadrant and the one-dimensional beam.