Answer to Question #165348 in Calculus for Vindhya

Question

double integral of xdydx +2ydzdx+ 3zdxdy over the surface of the spherex^2 +y^2 +z^2 is

Answer 1

Given integral is-

$\int \int xdxdy+\int \int 2ydzdx+\int \int3zdxdy$

Given Region is $x^2+y^2+z^2=1$

Let's evaluate the first term- $\int\int xdxdy$

Putting $z=0, \text { we get } x^2+y^2=1\implies y=\sqrt{1-x^2}$

$=\int_0^1\int_0^{\sqrt{1-x^2}} xdxdy$

$=\int_0^1x[y]_0^{\sqrt{1-x^2}}dx$

$=\int_0^1x(\sqrt{1-x^2})dx$

$\text{ Let }1-x^2=t$

$\\ -2xdx=dt\\ xdx=\dfrac{dt}{-2}$

$=\int_1^0\dfrac{\sqrt{t}}{-2}dt$

$=-\dfrac{1}{2}[\dfrac{t^\frac{3}{2}}{\frac{3}{2}}]_1^0$

$=\dfrac{1}{3}(1)=\dfrac{1}{3}~~~~~~~-(1)$

Solving integral of second term-

$\int\int_R 2ydxdz$

Putting the value of $y=\sqrt{1-x^2-z^2}$ in above integral as

$=2\int_0^1\int_0^{\sqrt{1-z^2}}\sqrt{1-z^2-x^2}dzdx$

$=2\int_0^1(\dfrac{x}{2}\sqrt{1-z^2-x^2}+\dfrac{1-z^2}{2}sin^{-1}(\dfrac{x}{\sqrt{1-z^2}})_0^{\sqrt{1-z^2}})dz$

$=2\int_0^1(\dfrac{1-z^2}{2}sin^{-1}(\dfrac{\sqrt{1-z^2}}{\sqrt{1-z^2}})-0)dz$

$=2\int_0^1\dfrac{1-z^2}{2}sin^{-1}(1)dz$

$=\int_0^1(1-z^2)\dfrac{\pi}{2}dz$

$=\dfrac{\pi}{2}(z-\dfrac{z^3}{3})_0^1$

$=\dfrac{\pi}{2}(1-\dfrac{1}{3})=\dfrac{\pi}{2}(\dfrac{2}{3})=\dfrac{\pi}{3}~~~~~~~~~-(2)$

Now integrate the third term-

$\int\int_R3zdxdy$

Now Putting the value $z =\sqrt{1-x^2-y^2}$ and applying the same procedure as above-

$=3\int_0^1\int_0^{\sqrt{1-y^2}}\sqrt{1-y^2-x^2}dxdy$

$=3\int_0^1(\dfrac{x}{2}\sqrt{1-y^2-x^2}+\dfrac{1-y^2}{2}sin^{-1}(\dfrac{x}{\sqrt{1-y^2}})_0^{\sqrt{1-y^2}})dy$

$=3\int_0^1(\dfrac{1-y^2}{2}sin^{-1}(\dfrac{\sqrt{1-y^2}}{\sqrt{1-y^2}})-0)dy$

$=3\int_0^1\dfrac{1-y^2}{2}sin^{-1}(1)dy$

$=\dfrac{3}{2}\int_0^1(1-y^2)\dfrac{\pi}{2}dy$

$=\dfrac{3\pi}{4}(y-\dfrac{y^3}{3})_0^1$

$=\dfrac{3\pi}{4}(1-\dfrac{1}{3})=\dfrac{3\pi}{4}(\dfrac{2}{3})=\dfrac{\pi}{2}~~~~~~~~~-(3)$

Adding the three equation we get-

$\int \int xdxdy+\int \int 2ydzdx+\int \int3zdxdy=\dfrac{1}{3}+\dfrac{\pi}{3}+\dfrac{\pi}{2}=\dfrac{5\pi}{6}+\dfrac{1}{3}$