Answer to Question #176845 in Calculus for Zulfikor

Question

a solid of constant density is bounded below by the plane z = 0 , on the sides by the ellptical cylinder x^2 + 4y^2 = 4, and above by the plane z = 2 - x . calculate a) volume b) center of mass c) moment of inertial about x-axis , y-axis and z-axis

Answer 1

a) The volume

$V=\iiint dxdydz$

Now, $z=0;x^2+4y^2=4;z=2-x$

Then, $\text{\textbraceleft}0\eqslantless z \eqslantless2-x, 0\eqslantless x \eqslantless \sqrt{4-4y^2}, 0\eqslantless y\eqslantless1 \text{\textbraceright}$

So the volume integral becomes

$V=\int_0^1 \int_0^{\sqrt{4-4y^2}}\int_0^{2-x}dzdxdy$

$V=\int_0^1 \int_0^{\sqrt{4-4y^2}}(2-x)dxdy$

$V=\int_0^1 [2x- \frac{x^2}{2}]_0^{\sqrt{4-4y^2}}dy$

$V=2[ \frac{y^3}{3}+y \sqrt{1-y^2}-y+arcsin (y)]_0^1$

$V=\pi- \frac{4}{3}$

b) Center of mass 

$M=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}\int_0^{2-x}dzdydx$

$M=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}[2-x]dydx$

$M=\int_{-2}^2 ((2-x)\sqrt{4-4y^2})dx$

$M=\int_{0}^2 2\sqrt{4-4y^2}-x\sqrt{4-4y^2})dx=4 \pi$

$M_{yz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}\int_0^{2-x}xdzdydx$

$M_{yz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}x[2-x]dydx$

$M_{yz}=\int_{-2}^2 ((2x-x^2)\sqrt{4-4y^2})dx$

$M_{yz}=-2\int_{0}^2 x^2\sqrt{4-4x^2} dx$

Let $x= 2sin \theta; dx=2cos \theta d\theta$

$M_{yz}=-32 \int_0 ^{\frac{\pi}{2}}sin^2 \theta cos^2 \theta d \theta =-2 \pi$

$M_{xz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}\int_0^{2-x}ydzdydx$

$M_{xz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}y[2-x]dydx$

$M_{xz}=\frac{2-x}{2}[\frac{(\sqrt{4-4y^2})^2}{2}-\frac{(\sqrt{4-4y^2})^2}{2}] =0$

So, the center of mass is

$\bar{x}= \frac{M_{yz}}{M}= \frac{-2 \pi}{4 \pi}= -\frac{1}{2}$

$\bar{y}= \frac{M_{xz}}{M}= \frac{0}{4 \pi}= 0$

c) Moment of inertial about the x-axis, y-axis, and z-axis

$M_{yz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}\int_0^{2-x}xdzdydx$

$M_{yz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}x[2-x]dydx$

$M_{yz}=\int_{-2}^2 ((2x-x^2)\sqrt{4-4y^2})dx$

$M_{yz}=-2\int_{0}^2 x^2\sqrt{4-4x^2} dx$

Let $x= 2sin \theta; dx=2cos \theta d\theta$

$M_{yz}=-32 \int_0 ^{\frac{\pi}{2}}sin^2 \theta cos^2 \theta d \theta =-2 \pi$

$M_{xz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}\int_0^{2-x}ydzdydx$

$M_{xz}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}y[2-x]dydx$

$M_{xz}=\frac{2-x}{2}[\frac{(\sqrt{4-4y^2})^2}{2}-\frac{(\sqrt{4-4y^2})^2}{2}] =0$

$M_{xy}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}\int_0^{2-x}zdzdydx$

$M_{xy}=\int_{-2}^2 \int_{-0.5\sqrt{4-4y^2}}^{0.5\sqrt{4-4y^2}}z[2-x]dydx$

$M_{xy}=\int_{-2}^2 \sqrt{4-x^2})dx+\frac{1}{4}[x^2\int_{-2}^2 \sqrt{4-x^2})dx-\int_{-2}^2 \sqrt{4-x^2})dx]$

$M_{xy}=-2\int_{0}^2 x^2\sqrt{4-4x^2} dx$

$M_{xy}=2* \frac{\pi}{2}+4 \pi =5 \pi$