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?

Question

Christopher Agresta · Accepted Answer

The volume is #(81pi)/4 - 9 = 54.6173# (4dp) #unit^3#

The graphs of the plane #x+z=9# and the surface #x^2+y^2=9# are as follows:

# " " = 9sqrt(9-y^2)-1/2(9-y^2) #

And so our double integral now becomes:

# v= int_0^3 {9sqrt(9-y^2)-1/2(9-y^2) }\ dy #

And for this integral we can split into the two parts

# I_1 = int_0^3 9sqrt(9-y^2) \ dy# and # I_2 = int_0^3 -1/2(9-y^2) \ dy #

We can just evaluate the second integral to get:

# I_2 = -1/2[ 9y-1/3y^3 ]_0^3 #
# \ \ \ = (-1/2){(9)(3)-1/3(27) - 0} #
# \ \ \ = -9 #

And for the first integral we use the substitution #y=3sinu#, which gives the result:

# I_1 = 9 int_0^3 sqrt(9-y^2) \ dy #
# \ \ \ = 9 [ysqrt(9-y^2)/2 + 9/2 arcsin(y/3) ]_0^3 #
# \ \ \ = 9 {(0+9/2pi/2) - (0+0) } #
# \ \ \ = (81pi)/4 #

NOTE - You may 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=1/4pi(3^2) = (9pi)/4# which again gives #I_2=9A = (81pi)/4#.

Combining our results gives the total volume as:

# v= (81pi)/4 - 9#
# \ \ = (81pi)/4 - 9#
# \ \ = 54.617251 ... #

Ezekiel Alexander · Answer

undefined To find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder $x^2+y^2=9$, and the plane $x+z=9$, we first need to determine the limits of integration for $x$, $y$, and $z$.

Since the solid is in the first octant, the limits of integration for $x$ will be from $0$ to $3$ (the intersection of the cylinder and the plane $x+z=9$). For $y$, the limits will be from $0$ to $\sqrt{9-x^2}$ (the equation of the cylinder). For $z$, the limits will be from $0$ to $9-x$.

Now, we set up the triple integral to find the volume:
$$ V = \iiint dV = \int_{0}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{9-x} dz \, dy \, dx $$

Evaluating this triple integral will give us the volume of the solid in the first octant.