Vector Calculus – Exercises II

Exercise 5

Evaluate the surface integral of ${\vec{u}= (xy,x,x+y)}$ over the surface of ${S}$ defined by ${z=0}$ with ${0 \leq x \leq 1}$ , ${0 \leq y \leq 2}$ , with the normal ${\vec{n}}$ directed in the positive ${z}$ direction.

Since the surface is ${z=0}$ the normal is ${\vec{n}=(0, 0, 1)}$ . Hence

$\displaystyle \vec{u}\cdot \vec{n}= x+y$

Thus the integral is

${\begin{aligned} \iint _S \vec{u}\cdot \vec{n} dS &= \int _0 ^1 \int _0 (x+y) dy dx \\ &= \int _0 ^1 [xy+y^2/2]_{y=0}^{y=2} dx \\ &= \int _0 ^1 (2x+2) dx \\ &= [x^2+2x]_{x=0}^{x=1} \\ &= 1+2 \\ &= 3 \end{aligned}}$

Exercise 6

Find the surface integral of ${\vec{u}\cdot \vec{n}}$ over ${S}$ , where ${S}$ is part of the surface ${z=x+y^2}$ with ${z<0}$ , ${x>-1}$ , ${\vec{u}=(2y+x,-1,0)}$ and ${\vec{n}}$ has a negative ${z}$ component.

${S}$ can be written as ${S:(x, y, x+y^2)}$

Hence the two vectors that are tangent to ${S}$ are

$\displaystyle \vec{a}= \frac{\partial}{\partial x}(x, y, x+y^2)=(1, 0, 1)$

and

$\displaystyle \vec{b}= \frac{\partial}{\partial y}(x, y, x+y^2)=(0, 1, 2y)$

and for ${\vec{n}dS}$ it is

$\displaystyle \vec{n}dS = \vec{a}\times\vec{b}= (1, 0, 1)\times (0, 1, 2y) = (-1,-2y, 1)$

Since by hypothesis the normal needs to have a negative ${z}$ component we will take the symmetric of the previous vector

$\displaystyle (1,2y, -1)$

So the integral is

${\begin{aligned} \iint_S \vec{u} \cdot \vec{n}dS &= \iint_S (2y+x,-1,0)\cdot (1,2y, -1) dS \\ &= \iint_S (2y+x-2y)dS \\ &= \iint_S x dS \end{aligned}}$

The region of integration is ${x+y^2<0}$ and ${x>-1}$ . So if we choose to integrate in ${x}$ first it is ${/1<x<-y^2}$ and for ${y}$ it is ${-1<y<1}$ .

Hence the previous integral is

${\begin{aligned} \iint_S x dS &= \int_{-1}^1 \int_{-1}^{-y^2} x dx dy \\ &= \int_{-1}^1 [x^2/2]_{-1}^{-y^2} dy \\ &=\dfrac{1}{2}\int_{-1}^1(y^4-1)dy \\ &= \dfrac{1}{2} \left( [y^5/5]_{-1}^1-2\right) \\ &= \dfrac{1}{2}\left( \dfrac{1}{5} + \dfrac{1}{5} /2 \right) \\ &= \dfrac{1}{5} -1 \\ &= -\dfrac{4}{5} \end{aligned}}$

Exercise 7

Find the volume integral of the scalar field ${x^2+y^2+z^2}$ over the region ${V}$ defined by ${0 \leq x \leq 1}$ , ${1 \leq y \leq 2}$ and ${0 \leq z \leq 3}$ .

${\begin{aligned} \int_{0}^3 \int_{1}^2 \int_{0}^1 (x^2+y^2+z^2)dxdydz &= \int_{0}^3 \int_{1}^2 [x^3+x(y^2+x^2)]_{x=0}^{x=1}dydz \\ &= \int_{0}^3 \int_{1}^2 (1/3+y^2+z^2)dydz \\ &=\int_{0}^3 \left[ y^3/3+y(1/3+z^2)\right]_{y=1}^{y=2} dz \\ &= \int_{0}^3 (7/3+1/3+z^2)dz \\ &= \int_{0}^3 (8/3+z^2)dz \\ &= 8+[z^3/3]_{z=0}^{z=3} \\ &= 8+9 \\ &= 17 \end{aligned}}$

Exercise 8

Find the volume of the section of the cylinder ${x^2+y^2=1}$ between the planes ${z=x+1}$ and ${z=-x-1}$ .

We know that for ${z}$ it is ${-x-1 < z < x+1}$ . For ${y}$ it is ${-\sqrt{1-x^2} < y < \sqrt{1-x^2} }$ and for ${x}$ it is ${-1 < x < 1 }$ .

Thus the integral is

${\begin{aligned} \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{-\sqrt{1-x^2}} \int_{-x-1}^{x+1} dzdydx &= \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{-\sqrt{1-x^2}} (2x+2) dydx \\ &= \int_{-1}^1 4(x+1)\sqrt{1-x^2} dx \\ &=\int_{-1}^1 4x\sqrt{1-x^2}dx + \int_{-1}^1 4\sqrt{1-x^2} dx \end{aligned}}$

The first integral vanishes because it is the integral of an odd function between symmetric limits.

For the second integral we’ll use a change of variable ${x=\sin\theta}$ , then ${dx=\cos\theta d\theta}$ . Hence for the second integral it is

${\begin{aligned} \int_{-1}^1 4\sqrt{1-x^2} dx &= 4\int_{-\pi /2}^{\pi /2} \cos ^2 \theta d\theta \\ &= 4\int_{-\pi /2}^{\pi /2}\dfrac{1+\cos(2\theta)}{2} d\theta \\ &= 2 \left(\int_{-\pi /2}^{\pi /2} d\theta + \int_{-\pi /2}^{\pi /2} \dfrac{\cos(2\theta)}{2} d\theta \right) dx \\ &= 2\pi + \dfrac{1}{2}\left[\dfrac{\sin (2\theta)}{2}\right]_{-\pi /2}^{\pi /2} \\ &= 2\pi + 0 \\ &= 2\pi \end{aligned}}$

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