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24. (Accuracy) Let $S$ be the funnel-shaped surface defined by $x^2 + y^2 = z^2$ for $1 \le z \le 3$ and $x^2 + y^2 = 1$ for $0 \le z \le 1$. (a) Sketch a cross section of the surface in the $xz$ plane. (b) Evaluate $\iint_S \langle -y, x, z \rangle \cdot \mathbf{n} \, dS$ if $S$ has normals oriented away from the inside of the funnel. Hint: Split the surface integral into two surfaces $S_1$ and $S_2$ where $S_1$ is the surface for $0 \le z \le 1$ and $S_2$ the remaining surface. Then use the fact that $\iint_S \langle -y, x, z \rangle \cdot \mathbf{n} \, dS = \iint_{S_1} \langle -y, x, z \rangle \cdot \mathbf{n} \, dS + \iint_{S_2} \langle -y, x, z \rangle \cdot \mathbf{n} \, dS.$

          24. (Accuracy) Let $S$ be the funnel-shaped surface defined by $x^2 + y^2 = z^2$ for $1 \le z \le 3$ and $x^2 + y^2 = 1$ for $0 \le z \le 1$.
(a) Sketch a cross section of the surface in the $xz$ plane.
(b) Evaluate
$\iint_S \langle -y, x, z \rangle \cdot \mathbf{n} \, dS$
if $S$ has normals oriented away from the inside of the funnel.
Hint: Split the surface integral into two surfaces $S_1$ and $S_2$ where $S_1$ is the surface for $0 \le z \le 1$ and $S_2$ the remaining surface. Then use the fact that
$\iint_S \langle -y, x, z \rangle \cdot \mathbf{n} \, dS = \iint_{S_1} \langle -y, x, z \rangle \cdot \mathbf{n} \, dS + \iint_{S_2} \langle -y, x, z \rangle \cdot \mathbf{n} \, dS.$

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24. (Accuracy) Let S be the funnel-shaped surface defined by x^2 + y^2 = z^2 for 1 ≤ z ≤ 3 and x^2 + y^2 = 1 for 0 ≤ z ≤ 1.
(a) Sketch a cross section of the surface in the xz plane.
(b) Evaluate
⟨ -y, x, z ⟩·𝐧 dS
if S has normals oriented away from the inside of the funnel.
Hint: Split the surface integral into two surfaces S1 and S2 where S1 is the surface for 0 ≤ z ≤ 1 and S2 the remaining surface. Then use the fact that
⟨ -y, x, z ⟩·𝐧 dS = ∬S1⟨ -y, x, z ⟩·𝐧 dS + ∬S2⟨ -y, x, z ⟩·𝐧 dS.

Added by Ray S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/05/2023

Step 1

To sketch the cross section of the surface in the xz plane, we need to eliminate the variable y from the equation. From the given equations, we have x + y = √z for 1 < z < 3 and x + y = 1 for 0 < z < 1. For the first equation, x + y = √z, we can solve for y to Show more…

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Texts: Solve ALL of the following, showing step by step: 24. Accuracy: Let S be the funnel-shaped surface defined by x + y = √z for 1 < z < 3 and x + y = 1 for 0 < z < 1. (a) Sketch a cross section of the surface in the xz plane. (b) Evaluate ∫∫S y,x,z dS if S has normals oriented away from the inside of the funnel. Hint: Split the surface integral into two surfaces S1 and S2, where S1 is the surface for 0 < z < 1 and S2 is the remaining surface. Then use the fact that ∫∫S -y,x,z dS = ∫∫S -y,x,2 dS.

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00:01 The question consider double integral over s z plus 3ui minus x square dx where s is z equal to 2 minus 3ui plus x square and this lies above triangle.

00:23 Triangle coordinates are 0 ,0 2 ,0 and 2 minus 4.

00:33 This can be represented as in the graph 0 ,0 this is 2 ,0 and this is 2 minus 4.

00:40 This is a required triangle.

00:42 So now the region is region is that x equals to 2 and y equals minus 2x.

00:51 So now ds equals root of zx square plus zy square plus y.

01:00 Now we know that z equal to minus 3y plus x square.

01:04 This implies that z of x equal 2x and zy equals minus 3.

01:12 So ds equal 2x whole square plus minus 3 whole square plus 1.

01:21 This gives 4x square plus minus 1 gives 4x square plus 10.

01:30 Now the required integral that is s z plus 3y minus x square ds equals region that is 0 to 2 and this is minus 2a, 0.

01:46 Z plus 3y minus x square into ds is 4x square plus 10 into dy dx.

01:56 So this equals 0 to 2, 2x minus 2x to 0 and this can be written as 2.

02:04 So the function was this.

02:07 This is 4x square plus 10 dy by dx.

02:10 Now how we are writing this? that is 2 minus 3y plus x square equals to z.

02:18 So from here we can write this as 2.

02:22 Now taking common 0 to 2 minus 2x.

02:27 We are taking 2 outside.

02:28 This will give 2...

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