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
\[ y^2 = 4ax \longrightarrow (1) \]
\[ x^2 = 4ay \longrightarrow (2) \]
First we will trace the curve and find points of intersection
\[ y^2 = 4ax \Rightarrow x = \frac{y^2}{4a} \]
\[ \left(\frac{y^2}{4a}\right)^2 = 4ay \]
\[ y^4 = 64a^3y \]
\[ y^4 - 64a^3y = 0 \Rightarrow y(y^3 - 64a^3) = 0 \]
\[ y = 0, y = 4a \]
\[ \text{For } y = 0 \Rightarrow x = 0 \]
\[ \text{For } y = 4a \Rightarrow x = 4a \]
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
\[ = \int^{4a}_0 dy \int^{\sqrt{4ay}}_{\large\frac{y^2}{4a}} dx \]
\[ = \int^{4a}_0 dy [x]^{\sqrt{4ay}}_{\large\frac{y^2}{4a}} \]
\[ = \int^{4a}_0 dy \left[\sqrt{4ay} - \frac{y^2}{4a} \right] \]
\[ = \left[\sqrt{4a}\frac{y^{3/2}}{\frac{3}{2}} - \frac{y^3}{12a} \right]^{4a}_0 \]
\[ = \left[\frac{4\sqrt{a}}{3}(4a)^{3/2} - \frac{(4a)^3}{12a} \right] \]
\[ = \frac{16}{3}a^2 \]
Center of Gravity
Center of gravity of a surface is given by: \[ \overline{x} = \frac{\iint \rho\hspace{5pt} x\hspace{5pt} dxdy}{\iint \rho\hspace{5pt} dxdy} \] \[ \overline{y} = \frac{\iint \rho\hspace{5pt} y\hspace{5pt} dxdy}{\iint \rho\hspace{5pt} dxdy} \]
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, \(\overline{x} = 0 \)
As it is given that its density varies as the square of the distance from the diameter, therefore
\[ \rho \propto y^2 \]
\[ \Rightarrow \rho = \alpha y^2 \]
\[ \overline{y} = \frac{\iint \rho\hspace{5pt} y\hspace{5pt} dxdy}{\iint \rho\hspace{5pt} dxdy} \]
\[ = \frac{\iint (\alpha y^2)\hspace{5pt} y\hspace{5pt}dxdy }{\iint (\alpha y^2)\hspace{5pt} dxdy} \]
\[ = \large\frac{\int^a_{-a} dx \int^{\sqrt{a^2-x^2}}_0 y^3dy }{\int^a_{-a} dx \int^{\sqrt{(a^2-x^2)}}_0 y^2dy}
\]
\[ = \large\frac{\int^a_{-a} dx \left[\frac{y^4}{4} \right]^{\sqrt{a^2-x^2}}_0 }{\int^a_{-a} dx
\left[\frac{y^3}{3} \right]^{\sqrt{a^2-x^2}}_0} \]
\[ = \frac{3\int^a_{-a} (a^2-x^2)^2dx }{4\int^a_{-a} (a^2-x^2)^{3/2}dx } \]
Put \(x = a\sin{\theta} \)
\[ = \frac{3\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(a^2 - a^2\sin^2{\theta})^2 a\cos{\theta}d\theta
}{4\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(a^2 - a^2\sin^2{\theta})^2 a\cos{\theta}d\theta} \]
\[ = \frac{3a}{4}\large\frac{\frac{4\times 2}{5\times 3}}{\frac{3\times 1}{4\times 2}\frac{\pi}{2}} \]
\[ = \left(\frac{3a}{4} \right)\left(\frac{8}{15} \right)\left(\frac{16}{3\pi} \right) \]
\[ = \frac{32a}{15\pi} \]
Hence Center of Gravity (0. \(32a/15\pi\))
