Evaluate the triple integral $\iiint _ { E } z d V$ , where E is the wedge in the first octant bounded by $y ^ { 2 } + z ^ { 2 } = 1$ , y = x, and the yz-plane.
A) $\frac { 1 } { 4 }$

B) $\frac { 1 } { 8 }$
C) $\frac { 1 } { 2 }$
D) $\frac { 2 } { 3 }$
E) $\frac { 3 } { 4 }$
F)1
G) $\frac { 5 } { 4 }$

H) $\frac { 4 } { 3 }$

Question

Evaluate the triple integral $\iiint _ { E } z d V$ , where E is the wedge in the first octant bounded by $y ^ { 2 } + z ^ { 2 } = 1$ , y = x, and the yz-plane.
A) $\frac { 1 } { 4 }$

B) $\frac { 1 } { 8 }$ 
C) $\frac { 1 } { 2 }$
D) $\frac { 2 } { 3 }$
E) $\frac { 3 } { 4 }$ 
F)1 
G) $\frac { 5 } { 4 }$
 
H) $\frac { 4 } { 3 }$

Quizplus · Accepted Answer

The region E is bounded by $y^2 + z^2 = 1$ in the yz-plane, which is a circle of radius 1. Since it's in the first octant, $0 \leq y \leq 1$ and $0 \leq z \leq \sqrt{1-y^2}$. The boundary $y = x$ means $0 \leq x \leq y$. Thus, the integral becomes $\int_0^1 \int_0^y \int_0^{\sqrt{1-y^2}} z \, dz \, dx \, dy$. Integrating $z$ with respect to $z$ from $0$ to $\sqrt{1-y^2}$ gives $\frac{1}{2}(1-y^2)$. Then, integrating with respect to $x$ from $0$ to $y$ simply multiplies this by $y$, and finally integrating with respect to $y$ from $0$ to $1$ gives the result $\frac{1}{8}$.

Evaluate the Triple Integral ∭EzdV\iiint _ { E } z d V∭E​zdV

Evaluate the Triple Integral $\iiint _ { E } z d V$