Applications of Multivariable Calculus Engineering Maths, Btech first year

Applications of Multivariable Calculus

Here we will study applications of Double Integral in Area and Center of Gravity.

Area : Sample Problem

Find the area bounded by the parabolas $y^2=4ax$ and $x^2=4ay$.

Solution

$$y^2=4ax⟶(1)$$ $$x^2=4ay⟶(2)$$ First we will trace the curve and find points of intersection $$y^2=4ax\Rightarrow x=\frac{y^2}{4a}$$ $${\left(\frac{y^2}{4a}\right)}^2=4ay$$ $$y^4=64a^{3}y$$ $$y^4-64a^{3}y=0\Rightarrow y(y^3-64a^3)=0$$ $$y=0,y=4a$$ $$\text{For }y=0\Rightarrow x=0$$ $$\text{For }y=4a\Rightarrow x=4a$$ So the points of intersection are (0, 0) and (4a, 4a)

We will be taking horizontal strip for calculation. So the limits for $x$ are $$x=\frac{y^2}{4a}\text{ to }x=\sqrt{4ay}$$ Limits for $y$ are $y=0\text{ to }y=4a$
So the required area is given by $$={\int }_0^{4a}dy{\int }_{\frac{y^2}{4a}}^{\sqrt{4ay}}dx$$ $$={\int }_0^{4a}dy[x{]}_{\frac{y^2}{4a}}^{\sqrt{4ay}}$$ $$={\int }_0^{4a}dy[\sqrt{4ay}-\frac{y^2}{4a}]$$ $$={[\sqrt{4a}\frac{y^{3/2}}{\frac{3}{2}}-\frac{y^3}{12a}]}_0^{4a}$$ $$=[\frac{4\sqrt{a}}{3}(4a{)}^{3/2}-\frac{(4a{)}^3}{12a}]$$ $$=\frac{16}{3}{a}^2$$

Center of Gravity

Center of gravity of a surface is given by: $$\bar{x}=\frac{\iint \rho xdxdy}{\iint \rho dxdy}$$ $$\bar{y}=\frac{\iint \rho ydxdy}{\iint \rho dxdy}$$

Sample Problem

Find the position of center of gravity of a semi-circulur lamina of radius of $a$ if its density varies as the square of the distance from the diameter.

Solution

The equation of circle is $x^2+y^2+z^2$
Let the diameter of the semi-circle is x-axis and the line perpendicular to diameter is y-axis.

By symmetry, $\bar{x}=0$
As it is given that its density varies as the square of the distance from the diameter, therefore $$\rho \propto y^2$$ $$\Rightarrow \rho =\alpha y^2$$ $$\bar{y}=\frac{\iint \rho ydxdy}{\iint \rho dxdy}$$ $$=\frac{\iint (\alpha y^2)ydxdy}{\iint (\alpha y^2)dxdy}$$ $$=\frac{{\int }_{-a}^{a}dx{\int }_0^{\sqrt{a^2-x^2}}{y}^{3}dy}{{\int }_{-a}^{a}dx{\int }_0^{\sqrt{(a^2-x^2)}}{y}^{2}dy}$$ $$=\frac{{\int }_{-a}^{a}dx{\left[\frac{y^4}{4}\right]}_0^{\sqrt{a^2-x^2}}}{{\int }_{-a}^{a}dx{\left[\frac{y^3}{3}\right]}_0^{\sqrt{a^2-x^2}}}$$ $$=\frac{3{\int }_{-a}^a(a^2-x^2{)}^{2}dx}{4{\int }_{-a}^a(a^2-x^2{)}^{3/2}dx}$$ Put $x=a\sin \theta$ $$=\frac{3{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(a^2-a^2{\sin }^2\theta {)}^{2}a\cos \theta d\theta }{4{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(a^2-a^2{\sin }^2\theta {)}^{2}a\cos \theta d\theta }$$ $$=\frac{3a}{4}\frac{\frac{4\times 2}{5\times 3}}{\frac{3\times 1}{4\times 2}\frac{\pi }{2}}$$ $$=\left(\frac{3a}{4}\right)\left(\frac{8}{15}\right)\left(\frac{16}{3\pi }\right)$$ $$=\frac{32a}{15\pi }$$ Hence Center of Gravity (0. $32a/15\pi$)