Evaluating a Surface Integral of a Sphere with radius t

Question

I had a question about a Surface Integral, I'm not sure if I'm just not understanding the notation, or if I'm right. The Surface Integral is: $$\iint_{\partial B(0,t)} \psi(\vec x) dS_{\vec x}$$ Where $$\psi(x) = \begin{cases} 1 & \text{ if } |\vec x|\leq 1 \\ 0 & \text{ otherwise}\end{cases}$$ Now, is the value equal to $4\pi t^2$, as we are just finding the surface area of the sphere with radius $t$? I'm not sure about the inclusion of $\psi$, but it's just stating that all vectors with magnitude smaller or equal to $1$ will give $1$, which wouldn't affect the outcome of the integral no?

Answer 1

I believe the answer to your above integral to be $$ 4\pi t^2 $$ for $t\leq 1$ and $0$ otherwise. As you said, if $t\leq 1$, you are just evaluating $$ \int\int_{\partial B(0,t)}dS_\vec{x}=4\pi t^2 $$ and integrating over $0$ otherwise.