MATH 2010 Exam 2 Review Problems

To evaluate the double integral $\iint_D xy \, dA$, we first need to determine the region of integration $D$ and set up the iterated integral.

1. Determine the Region $D$

The region $D$ is bounded by the curves $y = x$ and $y = \sqrt{x}$. To find the intersection points of these curves, we set the equations equal to each other:

Squaring both sides gives:

Thus, the intersection points occur at $x = 0$ and $x = 1$. The corresponding $y$-values are $y=0$ and $y=1$, giving the points $(0,0)$ and $(1,1)$.

On the interval $0 < x < 1$, we have $\sqrt{x} > x$. Therefore, the region $D$ can be described as:

2. Set Up the Integral

We can express the double integral as an iterated integral integrating with respect to $y$ first, then $x$:

3. Evaluate the Inner Integral

First, we integrate with respect to $y$, treating $x$ as a constant:

Substituting the limits of integration:

4. Evaluate the Outer Integral

Now, we integrate the result with respect to $x$ from $0$ to $1$:

Evaluating at the boundaries:

Finding a common denominator:

Final Answer

To compute the integral $\iint_D (x^2+y^2) \, dA$ using polar coordinates, we proceed with the following steps.

1. Coordinate Transformation We convert from Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$ using the relations:

Consequently, the term $x^2 + y^2$ becomes $r^2$, and the area element $dA$ becomes $r \, dr \, d\theta$.

2. Determine the Region of Integration The region $D$ is bounded by the circles $x^2+y^2=4$ and $x^2+y^2=9$, with the constraints $x \ge 0$ and $y \ge 0$.

• The equations of the circles translate to $r^2 = 4 \implies r = 2$ and $r^2 = 9 \implies r = 3$. Thus, the radial bounds are $2 \le r \le 3$.
• The constraints $x \ge 0$ and $y \ge 0$ restrict the region to the first quadrant, which corresponds to $0 \le \theta \le \frac{\pi}{2}$.

So, the region $D$ in polar coordinates is defined as:

3. Set Up the Integral Substituting the integrand and the differential element into the double integral:

4. Evaluate the Integral First, we evaluate the inner integral with respect to $r$:

Next, we evaluate the outer integral with respect to $\theta$:

Final Answer

To calculate the integral of $f(x,y) = (x^2+y^2)^{-3/2}$ over the region $D$ using polar coordinates, we proceed as follows.

1. Understand the Region $D$

The region $D$ is defined by two conditions:

• $x^2+y^2 \le 16$: This is a disk of radius 4 centered at the origin.
• $x+y \ge 4$: This is the half-plane above the line $x+y=4$.

The region $D$ is the portion of the disk that lies above the line $x+y=4$.

2. Find Intersection Points

To find where the line $x+y=4$ intersects the circle $x^2+y^2=16$, we substitute $y = 4-x$ into the circle equation:

Thus, $x = 0$ or $x = 4$. The intersection points are $(0, 4)$ and $(4, 0)$.

3. Convert to Polar Coordinates

Using polar coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $x^2 + y^2 = r^2$
• $dA = r \, dr \, d\theta$

The integrand becomes:

4. Determine the Bounds

• Radial bounds: From the line $x+y=4$, we have $r(\cos\theta + \sin\theta) = 4$, so $r = \frac{4}{\cos\theta + \sin\theta}$. From the circle, $r = 4$. Thus, $\frac{4}{\cos\theta + \sin\theta} \le r \le 4$.
• Angular bounds: The intersection points $(4, 0)$ and $(0, 4)$ correspond to $\theta = 0$ and $\theta = \frac{\pi}{2}$ respectively. Thus, $0 \le \theta \le \frac{\pi}{2}$.

5. Set Up the Integral

6. Evaluate the Inner Integral

7. Evaluate the Outer Integral

Final Answer

To evaluate the triple integral $\iiint_B (xz + yz^2) \, dV$, where $B = [0,1] \times [2,4] \times [0,2]$, we set up the iterated integral based on the bounds of the rectangular box $B$.

1. Set Up the Integral

The region $B$ is defined by the inequalities:

We can write the triple integral as an iterated integral. The order of integration does not matter for a rectangular box, but we will integrate with respect to $z$, then $y$, then $x$:

2. Evaluate the Inner Integral (with respect to $z$)

First, we integrate with respect to $z$, treating $x$ and $y$ as constants:

Substituting the limits $z=2$ and $z=0$:

3. Evaluate the Middle Integral (with respect to $y$)

Next, we integrate the result with respect to $y$ from $2$ to $4$, treating $x$ as a constant:

Simplifying the term with $y^2$:

Evaluating at the boundaries:

4. Evaluate the Outer Integral (with respect to $x$)

Finally, we integrate with respect to $x$ from $0$ to $1$:

Evaluating at the boundaries:

Final Answer

To evaluate the triple integral $\iiint_W 66z \, dV$ over the region $W$, we first need to determine the limits of integration based on the given inequalities.

1. Determine the Region of Integration

The region $W$ is defined by:

• $0 \le x \le 1$
• $x^2 \le y \le 1$
• $x-y \le z \le x+y$

We can set up the iterated integral in the order $dz \, dy \, dx$:

2. Evaluate the Inner Integral (with respect to $z$)

First, we integrate with respect to $z$, treating $x$ and $y$ as constants:

Substituting the limits:

Using the identity $(a+b)^2 - (a-b)^2 = 4ab$, where $a=x$ and $b=y$:

3. Evaluate the Middle Integral (with respect to $y$)

Now, we integrate the result with respect to $y$ from $x^2$ to $1$, treating $x$ as a constant:

4. Evaluate the Outer Integral (with respect to $x$)

Finally, we integrate with respect to $x$ from $0$ to $1$:

Evaluating at the boundaries:

Finding a common denominator for the fractions:

Final Answer

Let the vector field be $\vec{F} = \langle P, Q, R \rangle$, where:

1. Calculate the Divergence

The divergence of $\vec{F}$ is given by the dot product of the del operator $\nabla$ and the vector field $\vec{F}$:

We compute the partial derivatives:

Adding these together, we get:

2. Calculate the Curl

The curl of $\vec{F}$ is given by the cross product of the del operator $\nabla$ and the vector field $\vec{F}$:

We compute the necessary partial derivatives:

• For the $\mathbf{i}$-component: $$\frac{\partial R}{\partial y} = 0, \quad \frac{\partial Q}{\partial z} = 3e^{2y+3z} \implies 0 - 3e^{2y+3z} = -3e^{2y+3z}$$
• For the $\mathbf{j}$-component: $$\frac{\partial P}{\partial z} = 0, \quad \frac{\partial R}{\partial x} = 2x \implies 0 - 2x = -2x$$
• For the $\mathbf{k}$-component: $$\frac{\partial Q}{\partial x} = 0, \quad \frac{\partial P}{\partial y} = x \implies 0 - x = -x$$

Thus, the curl is:

Final Answer

To evaluate the line integral $\int_C \sqrt{1+36xy} \, ds$, we need to parametrize the curve $C$, determine the arc length element $ds$, and then compute the definite integral.

1. Parametrize the Curve $C$

The curve $C$ is given by the function $y = 4x^3$ from the point $(0,0)$ to $(1,4)$. We can use $x$ as the parameter.
Let $x = t$. Then $y = 4t^3$.
The range for $x$ is from $0$ to $1$, so the parameter $t$ ranges from $0$ to $1$.

2. Calculate the Arc Length Element $ds$

The differential arc length $ds$ is given by:

First, compute the derivatives with respect to $t$:

Substitute these into the formula for $ds$:

3. Express the Integrand in Terms of $t$

The integrand is $f(x,y) = \sqrt{1+36xy}$. Substitute $x=t$ and $y=4t^3$:

4. Set Up and Evaluate the Integral

Now, substitute the integrand and $ds$ into the line integral:

The two square root terms multiply to form the expression inside the root:

Now, evaluate the definite integral:

Final Answer

To address the note first, we determine if the vector field $\vec{F}$ is conservative by calculating its curl.

1. Is $\vec{F}$ Conservative?

A vector field $\vec{F} = \langle P, Q, R \rangle$ is conservative if and only if $\text{curl}(\vec{F}) = \vec{0}$ (assuming the domain is simply connected).
Given $\vec{F} = \langle xy, 3, z^3 \rangle$, we have $P = xy$, $Q = 3$, and $R = z^3$.
The curl is given by:

Calculating the components:

• $\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - 0 = 0$
• $\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0 - 0 = 0$
• $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - x = -x$

Thus,

Since $\text{curl}(\vec{F}) \neq \vec{0}$ (specifically, the $z$-component is $-x$, which is not identically zero), the vector field is not conservative. This is why we cannot use the Fundamental Theorem of Line Integrals and must evaluate the integral directly.

2. Compute the Line Integral

We evaluate $\int_C \vec{F} \cdot d\vec{r}$ using the parameterization $\vec{r}(t) = \langle \cos t, \sin t, t \rangle$ for $0 \le t \le \pi$.

Step 1: Find $d\vec{r}$

Step 2: Evaluate $\vec{F}$ on the curve Substitute $x = \cos t$, $y = \sin t$, and $z = t$ into $\vec{F}$:

Step 3: Compute the dot product $\vec{F} \cdot d\vec{r}$

Step 4: Integrate with respect to $t$

We can split this into three integrals:

• First integral: Let $u = \sin t$, then $du = \cos t \, dt$. The limits change from $0 \to 0$ and $\pi \to 0$.

  $$\int_0^{\pi} -\sin^2 t \cos t \, dt = \left[ -\frac{\sin^3 t}{3} \right]_0^{\pi} = 0 - 0 = 0$$

• Second integral:

  $$\int_0^{\pi} 3 \cos t \, dt = 3 [\sin t]_0^{\pi} = 3(0 - 0) = 0$$

• Third integral:

  $$\int_0^{\pi} t^3 \, dt = \left[ \frac{t^4}{4} \right]_0^{\pi} = \frac{\pi^4}{4} - 0 = \frac{\pi^4}{4}$$

Summing these results:

Final Answer

To solve these integrals, we first need to parametrize the line segment $C$.

1. Parametrize the Curve $C$

The curve $C$ is the line segment from $P_0 = (-2, 1, 0)$ to $P_1 = (-1, 2, 1)$.
The direction vector is $\vec{v} = P_1 - P_0 = \langle -1 - (-2), 2 - 1, 1 - 0 \rangle = \langle 1, 1, 1 \rangle$.
We can parametrize the curve using $t$ where $0 \le t \le 1$:

Thus, the coordinates are:

The derivative of the position vector is:

The magnitude of the derivative is:

Therefore, the differential arc length is $ds = \sqrt{3} \, dt$, and the differential vector is $d\vec{r} = \langle 1, 1, 1 \rangle \, dt$.

a. Scalar Line Integral

We need to set up $\int_C f(x,y,z) \, ds$ where $f(x,y,z) = yz - x^2$.

Step 1: Substitute the parametrization into $f$

Step 2: Simplify the integrand

Step 3: Set up the integral

Step 4: Evaluate

b. Vector Line Integral

We need to set up $\int_C \vec{F} \cdot d\vec{r}$ where $\vec{F} = \langle 2x-y, 2z, y-z \rangle$.

Step 1: Substitute the parametrization into $\vec{F}$

Step 2: Simplify the components of $\vec{F}$

• $x$-component: $2(-2+t) - (1+t) = -4 + 2t - 1 - t = t - 5$
• $y$-component: $2t$
• $z$-component: $1 + t - t = 1$ So, $\vec{F}(\vec{r}(t)) = \langle t - 5, 2t, 1 \rangle$.

Step 3: Compute the dot product $\vec{F} \cdot \vec{r}'(t)$

Step 4: Set up the integral

Step 5: Evaluate

Final Answer

a. $\displaystyle \int_C (yz - x^2) \, ds = \int_0^1 \sqrt{3}(5t - 4) \, dt = -\frac{3\sqrt{3}}{2}$
b. $\displaystyle \int_C \vec{F} \cdot d\vec{r} = \int_0^1 (3t - 4) \, dt = -\frac{5}{2}$

a. Show that $\vec{F}$ is conservative and find a potential function

Let $\vec{F} = \langle P, Q \rangle$, where $P = 3+2xy$ and $Q = x^2-3y^2$.
For $\vec{F}$ to be conservative on a simply connected domain (like $\mathbb{R}^2$), the condition $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ must hold.

Calculate the partial derivatives:

Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, the vector field $\vec{F}$ is conservative.

To find the potential function $f(x,y)$ such that $\nabla f = \vec{F}$, we solve the system:

1. $\frac{\partial f}{\partial x} = 3+2xy$
2. $\frac{\partial f}{\partial y} = x^2-3y^2$

Integrate the first equation with respect to $x$:

where $g(y)$ is an arbitrary function of $y$.

Now, differentiate this expression with respect to $y$ and set it equal to $Q$:

Integrate with respect to $y$:

Thus, the potential function is: