Question

Use polar coordinates to fithe volume of the solid inside both the cylinder x^2 + y^2 = 4 and the ellipsoid 4x^2 + 4y^2 + z^2 = 64

Answer 1

Answer:

$V = 64\pi . [\frac{16}{3} - \sqrt{3} ]$  unit^3 .. ( or equivalent )

Step-by-step explanation:

Solution:-

- The following surfaces are given as follows:

                                $x^2 + y^2 = 4 \\\\x^2 + 4y^2 + z^2 = 64$

- We will first have to investigate the region that lies inside both the cylinder and the ellipsoid or the region common to both surface.

Step 1: Coordinate transformation ( Cartesian ( x,y,z ) -> Polar ( r,θ,z )

- To convert the cartesian coordinates to polar/cylindrical coordinate system we will take the help of conversion equation given below:

                                $x^2 + y^2 = r^2$

- Make the substitution of the above equation in the given equation of the cylinder as follows:

                               $r^2 = 4\\r = +/- 2$

- Make the substitution of the transformation equation into the ellipsoid equation as follows:

                              $4 ( x^2 + y^2 ) + z^2 = 64\\\\4.r^2 + z^2 = 64\\\\z^2 = 4 ( 16 - r^2 )\\\\z = +/- 2.\sqrt{16 - r^2}$

Step 2: Sketch/Plot the region of volume

- We can either sketch or plot the surfaces in the cartesian coordinate system. I have utilized " Geogebra " 3D graphing utility.

- The purpose of graphing/sketching the surfaces is to "visualize" the bounds of the volume that lies inside both of the surfaces. This will help us in step 3 to set-up limits of integration. Moreover, the sketches also help us to see whether the enclosed volume is symmetrical about any axis.

- The plot is given as an attachment. From the plot we see that the volume of integration lies both above and below x-y plane ( z = 0 ). This result can be seen from the polar equation of surfaces ( +/- ) obtained.

- The Volume is defined by a shape of a vessel. I.e " Cylinder with two hemispherical caps on the circular bases "

- We see that enclosed volume is symmetrical about ( x-y ) plane. Each section of volume ( above and below x-y plane ) is bounded by the planes:

                         Upper half Volume:

                          $z = 0 - lower\\\\z = 2\sqrt{16 - r^2} - upper$

                        Lower Half Volume:

                         $z = -2\sqrt{16-r^2} - lower\\\\z = 0 - upper$

- We will simplify our integral manipulation by using the above symmetry and consider the upper half volume and multiply the result by 2.

Note: We can also find quadrant symmetry of the volume defined by the circular projection of the volume on ( x-y plane ); however, the attempt would lead to higher number of computations/tedious calculations. This invalidates the purpose of using symmetry to simplify mathematical manipulations.  

Step 3: Set-up triple integral in the cylindrical coordinate system

- The general formulation for setting up triple integrals in cylindrical coordinate system depends on the order of integration.

- We will choose the following order in the direction of "easing symmetry" i.e: (dz.dr.dθ)  

                             $V = \int\limits^f_e\int\limits^d_c\int\limits^b_a {} \, dz.(r.dr).dQ$

- Define the limits:

                 a: z = 0 - > ( x-y plane, symmetry plane )

                 b: z = 2√(16 - r^2) - > ( upper surface of ellipsoid )

            * Multiply the integral ( dz ) by " x2 "

                 c: r = 0  ( symmetry axis )

                 d: r = +2 ( circle of radius 2 units - higher )

              * Multiply the integral ( r.dr ) by " x2 "

                 e: θ = 0 ( initial point of angle sweep )

                 f: θ = 2π ( final point of angle sweep )

- The integral formulation becomes:

                       $V = \int\limits^f_e 2*\int\limits^d_c2*\int\limits^b_a{} \, dz.(r.dr).dQ \\\\V = 4 \int\limits^f_e \int\limits^d_c\int\limits^b_a{} \, dz.(r.dr).dQ$

Step 4: Integral evaluation

- The last step is to perform the integration of the formulation derived in previous step.

                   

                        $V = 4 \int\limits^f_e \int\limits^d_c{2.r.\sqrt{16 - r^2} } \, .dr.dQ\\\\V = 4 \int\limits^f_e {-\frac{2}{3} . (16 - r^2)^\frac{3}{2} } \,|\Limits^2_0 . dQ\\\\V = -\frac{8}{3} \int\limits^f_e { [ 24\sqrt{3} - 64] } .dQ\\\\V = -\frac{8}{3}. [ 24\sqrt{3} - 64] . 2\pi \\\\V = 64\pi . [\frac{16}{3} - \sqrt{3} ]$Answer.

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