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In cylindrical coordinates, the paraboloid $z = 9 + x^2 + y^2$ has the equation $z = 9 + r^2$ and the cylinder $x^2 + y^2 = 5$ has the equation $r = \sqrt{5}$ Step 2 Therefore, the region E enclosed by the paraboloid, the cylinder, and the xy-plane is described by $E = \{(r, \theta, z) | 0 \le z \le 9 + r^2, 0 \le r \le \sqrt{5}, 0 \le \theta \le 2\pi\}$ Step 3 To evaluate $\iiint_E e^z dV = \int_0^{2\pi} \int_0^{\sqrt{5}} \int_0^{9+r^2} e^z r dz dr d\theta$, we first calculate the innermost integral. $\int_0^{9+r^2} e^z r dz = [re^z]_0^{9+r^2} = re^{9+r^2} - r$ Step 4 Next, we have $\int_0^{\sqrt{5}} (re^{9+r^2} - r) dr = \int_0^{\sqrt{5}} re^{9+r^2} dr - \int_0^{\sqrt{5}} r dr$ The first integral requires the substitution $u = r^2 + 9$ and $du = 2r dr$, which means that $r dr = \frac{1}{2} du$. Step 5 When $r = 0$, we have $u = 9$, and when $r = \sqrt{5}$, we have $u = 14$. Thus, $\int_0^{\sqrt{5}} re^{9+r^2} dr = \frac{1}{2} \int_9^{14} e^u du = \frac{1}{2} (e^{14} - e^9)$

          In cylindrical coordinates, the paraboloid $z = 9 + x^2 + y^2$ has the equation $z = 9 + r^2$ and the cylinder $x^2 + y^2 = 5$ has the equation $r = \sqrt{5}$

Step 2
Therefore, the region E enclosed by the paraboloid, the cylinder, and the xy-plane is described by
$E = \{(r, \theta, z) | 0 \le z \le 9 + r^2, 0 \le r \le \sqrt{5}, 0 \le \theta \le 2\pi\}$

Step 3
To evaluate $\iiint_E e^z dV = \int_0^{2\pi} \int_0^{\sqrt{5}} \int_0^{9+r^2} e^z r dz dr d\theta$, we first calculate the innermost integral.
$\int_0^{9+r^2} e^z r dz = [re^z]_0^{9+r^2} = re^{9+r^2} - r$

Step 4
Next, we have
$\int_0^{\sqrt{5}} (re^{9+r^2} - r) dr = \int_0^{\sqrt{5}} re^{9+r^2} dr - \int_0^{\sqrt{5}} r dr$
The first integral requires the substitution $u = r^2 + 9$ and $du = 2r dr$, which means that $r dr = \frac{1}{2} du$.

Step 5
When $r = 0$, we have $u = 9$, and when $r = \sqrt{5}$, we have $u = 14$. Thus,
$\int_0^{\sqrt{5}} re^{9+r^2} dr = \frac{1}{2} \int_9^{14} e^u du = \frac{1}{2} (e^{14} - e^9)$

$In cylindrical coordinates, the paraboloid z = 9 + x^2 + y^2 has the equation z = 9 + r^2 and the cylinder x^2 + y^2 = 5 has the equation r = √(5) Step 2 Therefore, the region E enclosed by the paraboloid, the cylinder, and the xy-plane is described by E = {(r, θ, z) | 0 ≤ z ≤ 9 + r^2, 0 ≤ r ≤√(5), 0 ≤θ≤ 2π} Step 3 To evaluate e^z dV = ∫0^2π∫0^√(5)∫0^9+r^2 e^z r dz dr dθ, we first calculate the innermost integral. ∫0^9+r^2 e^z r dz = [re^z]0^9+r^2 = re^9+r^2 - r Step 4 Next, we have ∫0^√(5) (re^9+r^2 - r) dr = ∫0^√(5) re^9+r^2 dr - ∫0^√(5) r dr The first integral requires the substitution u = r^2 + 9 and du = 2r dr, which means that r dr = (1)/(2) du. Step 5 When r = 0, we have u = 9, and when r = √(5), we have u = 14. Thus, ∫0^√(5) re^9+r^2 dr = (1)/(2)∫9^14 e^u du = (1)/(2) (e^14 - e^9)$

Added by Joshua M.

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