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$Evaluate the integral, where E is enclosed by the parabolold z=9+x^(2)+y^(2), the cyllinder x^(2)+y^(2)=5, and the xy-plane. Use cylindrical coordinates. ∭_(E)e^(2)dv Step 1 In cylindrical coordinates, the parabolold z=9+x^(2)+y^(2) has the equation z=9+r^(2),9+r^(2) and the cylinder x^(2)+y^(2)=5 has the equation r=sqrt5 Sim 2 Therefore, the reglon E enclosed by the parabolold, the cylinder, and the xy-plane is described by E={(r,theta,z)∣0 <= z <= ◻◻^(.0 <= r <= sqrt5),quad0 <= theta <= 2pi:}$

Evaluate the integral, where $E$ is enclosed by the parabolold $z=9+x^{2}+y^{2}$ , the cyllinder $x^{2}+y^{2}=5$ , and the $x y$ -plane. Use cylindrical coordinates. $\newline$ $\iiint_{E} e^{2} d v$ $\newline$ Step $1$ $\newline$ In cylindrical coordinates, the parabolold $z=9+x^{2}+y^{2}$ has the equation $z=9+r^{2}, 9+r^{2}$ and the cylinder $x^{2}+y^{2}=5$ has the equation $r=\sqrt{5}$ $\newline$ $\operatorname{Sim} 2$ $\newline$ Therefore, the reglon $E$ enclosed by the parabolold, the cylinder, and the $x y$ -plane is described by $\newline$ $E=\left\{(r, \theta, z) \mid 0 \leq z \leq \square \square^{.0 \leq r \leq \sqrt{5}}, \quad 0 \leq \theta \leq 2 \pi\right.$

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Find equations of tangent lines using limits

Full solution

Q. Evaluate the integral, where

E

is enclosed by the parabolold

z=9+x^{2}+y^{2}

, the cyllinder

x^{2}+y^{2}=5

, and the

x y

-plane. Use cylindrical coordinates.

\newline

\iiint_{E} e^{2} d v

\newline

Step

1

\newline

In cylindrical coordinates, the parabolold

z=9+x^{2}+y^{2}

has the equation

z=9+r^{2}, 9+r^{2}

and the cylinder

x^{2}+y^{2}=5

has the equation

r=\sqrt{5}

\newline

\operatorname{Sim} 2

\newline

Therefore, the reglon

E

enclosed by the parabolold, the cylinder, and the

x y

-plane is described by

\newline

E=\left\{(r, \theta, z) \mid 0 \leq z \leq \square \square^{.0 \leq r \leq \sqrt{5}}, \quad 0 \leq \theta \leq 2 \pi\right.

Calculate total tape needed: To find out how many rolls of tape the electrician needs, we divide the total amount of tape needed by the amount of tape on each roll. $8,000 \, \text{cm} \div 2,000 \, \text{cm}/\text{roll} = 4 \, \text{rolls}.$

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