Winter 2015 – Q.6

6. (a) Evaluate $latex \iint r^3dr\;d\theta,$ over the area bounded between the circles $latex r=2cos\theta$ and $latex r=4cos\theta.$ [6M]

Given that $latex r=2\cos\theta$ is a circle with center (0, 1) and radius 1 and $latex r=4\cos\theta$ is a circle with center (0, 2) and radius 2 as shown in figure the shaded area between these circle is the region of integration

$latex \begin{array}{l}\iint r^3dr\;d\theta\\\\=2\int_0^\frac{\mathrm\pi}2\int_{2\cos\theta}^{4\cos\theta}r^3dr\;d\theta\\\\=2\int_0^\frac{\mathrm\pi}2\left[\frac{r^4}4\right]_{2\cos\theta}^{4\cos\theta}d\theta\\\\=\frac24\int_0^\frac{\mathrm\pi}2\left[\left(4\cos\theta\right)^4-\left(2\cos\theta\right)^4\right]d\theta\\\\=\frac12\int_0^\frac{\mathrm\pi}2\left[256\cos^4\theta-16\cos^4\theta\right]d\theta\\\\=\frac12\times240\int_0^\frac{\mathrm\pi}2\cos^4\theta\;d\theta\\\\=120\times\frac34\times\frac12\times\frac{\mathrm\pi}2\\\\=\frac{45\mathrm\pi}2\\\\\therefore\iint r^3dr\;d\theta=\frac{45\mathrm\pi}2\end{array}$

(b) Find the mass of an elliptic plate $latex \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where the density at any point (x, y) on it is $latex rho=mu xy.$ [6M]

Let the required mass be M which is four time the mass in the firest quadrant for the region OAB, x varies from 0 to a and y varies from 0 to $latex \frac ba\sqrt{a^2-x^2}$

$latex \begin{array}{l}M=4\int\limits_R\int e\;dx\;dy\\\\\;\;\;=4\int_0^a\int_0^{\frac ba\sqrt{a^2-x^2}}u\;x\;y\;dy\;dx\\\\\;\;\;=4\int_0^aux\left[\frac{y^2}2\right]_0^{\frac ba\sqrt{a^2-x^2}}dx\\\\\;\;\;=2u\int_0^ax\frac{b^2}{a^2}\left(a^2-x^2\right)dx\\\\\;\;\;=2u\;\frac{b^2}{a^2}\int_0^a\left(a^2x-x^3\right)dx\\\\\;\;\;=2u\;\frac{b^2}{a^2}\left[\frac{a^2x^2}2-\frac{x^4}4\right]_0^a\\\\\;\;\;=2u\;\frac{b^2}{a^2}\left[\frac{a^4}2-\frac{a^4}4\right]\\\\\;\;\;=2u\;\frac{b^2}{a^2}\times\frac{a^4}4\\\\\;\;\;=\frac{ua^2b^2}2\\\\\therefore Mass\;M=\frac{ua^2b^2}2\end{array}$

We have

$latex \int_0^x\int_0^{1-x}\;\int_0^{1-x-y}xyz\;dz\;dy\;dx$

$latex \begin{array}{l}=\int_0^1\int_0^{1-x}xy\;\left[\int_0^{1-x-y}z\;dz\right]\;dy\;dx\\\\=\int_0^1\int_0^{1-x}xy\;\left[\frac{z^2}2\;\right]_0^{1-x-y}\;dy\;dx\\\\=\int_0^1\int_0^{1-x}\frac{xy\left(1-x-y\right)^2}2\;\;dy\;dx\\\\=\frac12\int_0^1\int_0^{1-x}xy\left(1+x^2+y^2-2x-2y+2xy\right)\;\;dy\;dx\\\\=\frac12\int_0^1\left[\int_0^{1-x}\left[xy+x^3y+xy^3-2x^2y-2xy^2+2x^2y^2\right]\;\;dy\right]\;dx\end{array}$

$latex \begin{array}{l}=\frac12\int_0^1\left[x\left\{\frac{y^2}2\right\}_0^{1-x}+x^3\left\{\frac{y^2}2\right\}_0^{1-x}+x\left\{\frac{y^4}4\right\}_0^{1-x}-2x^2\left\{\frac{y^2}2\right\}_0^{1-x}-2x\left\{\frac{y^3}3\right\}_0^{1-x}+2x^2\left\{\frac{y^3}3\right\}_0^{1-x}\;\right]\;dx\\\\=\frac1{24}\int_0^1\left[6x\left(1-2x+x^2\right)+6x^3\left(1-2x+x^2\right)+3x\left(1-x^4+4x^3+6x^2-4x\right)-12x^2\left(1-2x+x^2\right)-8x\left(1-x^3+3x^2-3x\right)+8x^2\left(1-x^3+3x^2-3x\right)\right]dx\\\\=\frac1{24}\int_0^1\left[6x-12x^2+6x^3+6x^3-12x^4+6x^5+3x+3x^5+-12x^4+18x^3-12x^2-12x^2+24x^3-12x^4-8x+8x^2-24x^3+3x^5-12x^4+18x^3-12x^2-12x^2+24x^2+8x^2-8x^5+24x^4-24x^3\right]dx\\\\=\frac1{24}\int_0^1\left[x^5-4x^4+6x^3-4x^2+x\right]dx\\\\=\frac1{24}\left[\frac{x^6}6-4\frac{x^5}5+6\frac{x^4}4-4\frac{x^3}3+\frac{x^2}2\right]_0^1\\\\=\frac1{24}\left[\frac16-\frac45+\frac64-\frac43+\frac12\right]\\\\=\frac1{24}\left[\frac{5-24+45-40-15}{30}\right]\\\\=\frac1{24}\times\frac1{30}\\\\=\frac1{720}\\\\\therefore\int_0^x\int_0^{1-x}\;\int_0^{1-x-y}xyz\;dz\;dy\;dx=\frac1{720}\end{array}$