(1) Let $D:=\{(x,y,z) \in \Bbb R^3 : z \ge x^2+y^2, z \le 1\}$. Find the surface area $S$ of $D$.
(2) Let $a,b \in \Bbb R$ and let $D:=\{(x,y,z) \in \Bbb R^3 : z \ge x^2+y^2, z \le 2ax+2by+1\}$. Find the volume $V$ of $D$.
I know that the surface area of the surface given by the equation $z=f(x,y)$ where $(x,y)$ is a point from the region $D$ in the $xy$-plane is given by $S=\int\int_D\sqrt{(f_x)^2+(f_y)^2+1}dA$. However, the domains given in the questions are given by inequalities that I'm not familiar with. Formulation for the volume seems not clear to me either.
Any help would be much appreciated.
As regards (1), $D$ is a truncated paraboloid and its total surface is equal to the area of the disc $\{(x,y,1) \in \Bbb R^3 : x^2+y^2\leq 1\}$, that is $\pi$ plus the "lateral" area given by the integral $$\iint_{x^2+y^2\leq 1}\sqrt{(f_x)^2+(f_y)^2+1}\ dx dy$$ where $z=f(x,y)=x^2+y^2$. Can you take it from here?
As regards (2), the volume of $D$ is the volume of another truncated paraboloid (this time the cutting plane is $z=2ax+2by+1$). The intersection tells us what domain of the $xy$-plane where we have to integrate: $$x^2+y^2=z=2ax+2by+1\implies (x-a)^2+(y-b)^2=1+a^2+b^2$$ that is a disc centred at $(a,b)$ of radius $\sqrt{1+a^2+b^2}$. Hence the volume is given by $$\iint_{(x-a)^2+(y-b)^2\leq 1+a^2+b^2}\int_{z=x^2+y^2}^{2ax+2by+1}dz\,dxdy\\ \iint_{(x-a)^2+(y-b)^2\leq 1+a^2+b^2}(2ax+2by+1-(x^2+y^2))\,dxdy \\ \iint_{(x-a)^2+(y-b)^2\leq 1+a^2+b^2}(a^2+b^2+1-(x-a)^2-(y-b)^2)\,dxdy \\ \iint_{X^2+Y^2\leq 1+a^2+b^2}(a^2+b^2+1-X^2-Y^2)\,dXdY \\ \int_{\theta=0}^{2\pi}\int_{\rho=0}^{\sqrt{1+a^2+b^2}}(a^2+b^2+1-\rho^2)\,\rho d\rho d\theta.$$ Can you take it from here?