Answer:
[tex]\int_Cy^2dx+z^2dy+(1-x^2)dz=\dfrac{1}{3}[/tex]
Step-by-step explanation:
To find: [tex]I=\int_Cy^2dx+z^2dy+(1-x^2)dz[/tex]
where, C is the quarter of the circle of radius 1 in the xz plane.
The center at origin.
[tex]x=\cos t[/tex]
[tex]z=\sin t[/tex]
where, [tex]0\leq t\leq \dfrac{\pi}{2}[/tex] because it is quarter circle.
The circle oriented counterclockwise when viewed from the positive y–axis.
Therefore, [tex]y=0[/tex]
If x = cos t, then dx = -sin t dt
If z = sin t, then dz = cos t dt
If y = 0 , then dy = 0
Substitute the value into integral.
[tex]I=\int_0^{\pi/2}(0^2(-\sin tdt)+(\sin^2t)(0dt)+(1-\cos^2t)(\cos tdt))[/tex]
[tex]I=\int_0^{\pi/2}\sin^2t\cos tdt[/tex]
Let [tex]\sin t=\theta [/tex]
[tex]\cos t dt=d\theta[/tex]
[tex]t\rightarrow 0, \theta \rightarrow 0[/tex]
[tex]t\rightarrow \dfrac{\pi}{2},\theta \rightarrow 1[/tex]
Change the integral limit and variable
[tex]I=\int_0^1\theta^2d\theta[/tex]
[tex]I=\dfrac{\theta^3}{3}|_0^1[/tex]
[tex]I=\dfrac{1}{3}(1^3-0^3)[/tex]
[tex]I=\dfrac{1}{3}[/tex]
Hence, the value of integral is [tex]\dfrac{1}{3}[/tex]
