Translating the integral we get that $$\int_{S_t(x)} y^2\:dA(y) = \int_{S_t(0)} (y+x)^2 \:dA(y)$$ Then use the fact that $dA(y) = t^2d\Omega$ $$\int_{S_t(0)} y^2+2y\cdot x + x^2 \:dA(y) = \int_{S^2} t^4 + x^2t^2\:d\Omega + \int_{S_t(0)} 2y\cdot x \:dA(y) = 4\pi t^2(x^2+t^2)$$ where the secon...