Hey I am new to this sort of problems and I want to check if my solution is correct:
Consider $P := \{x ∈\mathbb{R}^3 : x_3 = x_1^2 + x_2^2 < 4\}$ . Sketch $P$ and calculate $S_2(P)$
The set $P$ is the region in $\mathbb{R}^3$ that lies below the surface of the paraboloid $x_3=x_1^2+x_2^2$ and inside the cylinder $x_1^2+x_2^2=4$.
To sketch this, we can first draw the curve $x_1^2+x_2^2=4$ in the $x_1x_2$-plane. This is a circle of radius $2$ centered at the origin. We can then draw the surface $x_3=x_1^2+x_2^2$ above this circle. This is a paraboloid with its vertex at the origin and its axis along the $x_3$-axis. The intersection of the paraboloid with the cylinder is the part of the paraboloid that lies inside the cylinder and above the circle.
To calculate the surface area of $P$, we can use the formula $S_2(P) = \iint_D \sqrt{1 + (\frac{\partial z}{\partial x_1})^2 + (\frac{\partial z}{\partial x_2})^2} dA$, where $z=x_1^2+x_2^2$ and $D$ is the projection of $P$ onto the $x_1x_2$-plane, which is the circle $x_1^2+x_2^2=4$.
We have $\frac{\partial z}{\partial x_1} = 2x_1$ and $\frac{\partial z}{\partial x_2} = 2x_2$, so
$$S_2(P) = \iint_D \sqrt{1 + 4x_1^2 + 4x_2^2} dA = \int_0^{2\pi} \int_0^2 \sqrt{1 + 4r^2} r dr d\theta$$
This integral can be evaluated using the substitution $u=1+4r^2$, which gives
$$S_2(P) = \int_0^{2\pi} \int_1^9 \frac{1}{8}\sqrt{u} du d\theta = \frac{\pi}{2}(9\sqrt{2}-1)$$
Therefore, the surface area of $P$ is $\frac{\pi}{2}(9\sqrt{2}-1)$.
Am I right or am I doing some errors?
Up to the evaluation of the $r$ integral it is correct, but $u$ should range from $1$ to $17$ after the substitution, not $1$ to $9$: $$\int_1^{17}\frac18\sqrt u\,du=\frac18\cdot\frac23[u^{3/2}]_1^{17}=\frac1{12}(17^{3/2}-1)$$ So the correct answer is $\frac\pi6(17^{3/2}-1)$.