Evaluate the integral $I=\iint\limits_S fdS\ \ \text{where}\ f=1-x^2-y^2-z^2$

Question

Evaluate the following integral: $$ \begin{aligned} &I=\iint\limits_S fdS\ \ \text{where}\ f=1-x^2-y^2-z^2,\\ &S \text{ is a part of the plane $x+y+z=1$ that lies inside of the sphere } x^2+y^2+z^2=1 \end{aligned} $$

I tried to approach this problem using sphere parametrization: $$ dS=R^2\sin\theta\ d\theta,\ \ I=\int\limits_0^{2\pi}d\varphi\int\limits_0^\pi(1-R^2)R^2\sin\theta\ d\theta $$ Then I understood that I had to express $R$ in terms of $\varphi$ and $\theta$, and the last integral is actually not that simple. Maybe there is a better approach to this problem?

Answer 1

The intersection of the plane $x+y+z=1$ and $x^2+y^2+z^2=1$ is a circle of radius $r=\sqrt{\frac23}$. And, from symmetry

$$I=\iint\limits_S (1-x^2-y^2-z^2)dS= \iint\limits_S (1-3z^2)dS =\frac{2\pi}3 -3 \iint\limits_S z^2dS\tag1 $$

For convenience of integration, rotate the coordinates to have the $z$-axis aligned with the normal direction of the plane $x+y+z=1$, i.e. $(1,1,1)$, which simplifies the equation of the plane $w=\frac1{\sqrt3}$ and

$$z= \frac1{\sqrt3}w- \frac{\sqrt2}{\sqrt3}u $$ Then, the integral becomes

$$ \iint_S z^2dS =\iint_{u^2+v^2\le \frac23} (\frac1{\sqrt3} \frac1{\sqrt3}- \frac{\sqrt2}{\sqrt3}u)^2 dudv=\frac{4\pi}{27} $$

Plug into (1) to obtain the integral

$$I= \frac{2\pi}9$$

Answer 2

If you use sphere parametrization, $\theta$ is fixed. In this case, $\theta = \frac{\pi}{4}$ and $dS = R^2 sin(\frac{\pi}{4}) dR d\varphi$.