Evaluate $\iiint_E z^2 \, dV$, where $E$ is the solid hemisphere defined by the inequality $x^2 + y^2 + z^2 \leq 4$, with $z \geq 0$.**Explanation:**- This is a triple integral problem where the region $E$ is a solid hemisphere. - The equation $x^2 + y^2 + z^2 \leq 4$ represents a sphere of radius 2 centered at the origin.- The condition $z \geq 0$ indicates the upper hemisphere.- The goal is to compute the volume integral of $z^2$ over this region.

The given region E is x2+y2+z2≤4 and z≥0.Considerx2+y2+z2=4z=4-x2-y2Put z=0,thenx2+y2=4It is circle…

Answered: Evaluate 22 dV, where E is the solid…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Evaluate $\iiint_E z^2 \, dV$, where $E$ is the solid hemisphere defined by the inequality $x^2 + y^2 + z^2 \leq 4$, with $z \geq 0$.

**Explanation:**
- This is a triple integral problem where the region $E$ is a solid hemisphere.
- The equation $x^2 + y^2 + z^2 \leq 4$ represents a sphere of radius 2 centered at the origin.
- The condition $z \geq 0$ indicates the upper hemisphere.
- The goal is to compute the volume integral of $z^2$ over this region.

Expert Solution

Step 1

$\begin{array}{l} The given region E is x^{2} + y^{2} + z^{2} \leq 4 and z \geq 0 . \\ Consider \\ x^{2} + y^{2} + z^{2} = 4 \\ z = \sqrt{4 - x^{2} - y^{2}} \\ Put z = 0, then \\ x^{2} + y^{2} = 4 \\ It is circle of radius 2 units with centre at (0, 0) . \\ Put x = r \cos (θ) and y = r \sin (θ), then \\ d V = d x d y d z = r d r d θ d z \\ Now, the new region of integration is \\ R = \{(r, θ, z) / 0 \leq θ \leq 2 π, 0 \leq r \leq 2 and 0 \leq r \leq \sqrt{4 - x^{2} - y^{2}}\} . \end{array}$

Step 2

$\begin{array}{l} Now \\ \int_{}^{} \int_{}^{} \int_{E}^{} z^{2} d V = \int_{0}^{2 π} \int_{0}^{2} \int_{0}^{\sqrt{4 - r^{2}}} z^{2} r d r d θ d z \\ = \int_{0}^{2 π} {\int_{0}^{2} r (\frac{z^{3}}{3})}_{0}^{\sqrt{4 - r^{2}}} d r d θ \\ = \int_{0}^{2 π} \int_{0}^{2} r (\frac{{(4 - r^{2})}^{\frac{3}{2}}}{3}) d r d θ \\ = \frac{1}{3} \int_{0}^{2 π} \int_{0}^{2} r ({(4 - r^{2})}^{\frac{3}{2}}) d r d θ \\ Put 4 - r^{2} = t \Rightarrow r d r = - \frac{d t}{2} \end{array}$

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