NEOCODE

Double integrals, change of order of integration, change of variables

1. The value of \(\int_{0}^{1} \int_{0}^{2} (x + y) \,dx\,dy\) is:

Correct Answer: c) 3

Correct Answer: c) 3

Explanation:

We are tasked with evaluating the double integral:

\[ \int_0^1 \int_0^2 (x + y) \, dx \, dy \] Step 1: First, integrate with respect to \(x\): \[ \int_0^2 (x + y) \, dx = \left[ \frac{x^2}{2} + yx \right]_0^2 \] \[ = \left( \frac{4}{2} + 2y \right) - (0 + 0) = 2 + 2y \] Step 2: Now, integrate with respect to \(y\): \[ \int_0^1 (2 + 2y) \, dy = \left[ 2y + y^2 \right]_0^1 \] \[ = (2 \cdot 1 + 1^2) - (2 \cdot 0 + 0^2) = 2 + 1 = 3 \] Therefore, the value of the integral is **3**.

2. When reversing the order of integration for \(\int_{0}^{1} \int_{y^2}^{1} f(x,y) \,dx\,dy\), the new limits are:

Correct Answer: a) \(\int_{0}^{1} \int_{0}^{\sqrt{x}} f(x,y) \,dy\,dx\)

Explanation:

We are given:

\[ \int_0^1 \int_{y^2}^1 f(x,y) \, dx \, dy \] Step 1: Identify the bounds for \(x\) and \(y\). - The outer integral has \(y\) from **0 to 1**. - The inner integral has \(x\) from \(y^2\) to **1**. This means \(y^2 \le x \le 1\). Step 2: Rearrange the bounds for reversing the order of integration. - \(x\) varies from **0 to 1**. - For a given \(x\), \(y^2 \le x \rightarrow y \le \sqrt{x}\). - Since \(y \ge 0\), we get \(0 \le y \le \sqrt{x}\). Therefore, the reversed integral is: \[ \int_0^1 \int_0^{\sqrt{x}} f(x,y) \, dy \, dx \]

3. The double integral \(\iint_{D} (x^2 + y^2) \,dA\) over the unit circle \(D\) becomes in polar coordinates:

Correct Answer: b) \(\int_{0}^{2\pi} \int_{0}^{1} r^3 \,dr\,d\theta\)

Explanation:

In polar coordinates:

  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( dA = r \, dr \, d\theta \)
The given integral becomes: \[ x^2 + y^2 = r^2 (\cos^2 \theta + \sin^2 \theta) = r^2 \] Therefore: \[ \iint_D (x^2 + y^2) \, dA = \int_0^{2\pi} \int_0^1 r^2 \cdot r \, dr \, d\theta \] Which simplifies to: \[ \int_0^{2\pi} \int_0^1 r^3 \, dr \, d\theta \] \]

4. For the transformation \(u = x + y\), \(v = x - y\), the Jacobian determinant \(\frac{\partial(x,y)}{\partial(u,v)}\) is:

Correct Answer: b) -1/2

Explanation:

To find the Jacobian determinant for the given transformation:

\( u = x + y, \quad v = x - y \)

Step 1: Express \(x\) and \(y\) in terms of \(u\) and \(v\)

We can solve for \(x\) and \(y\) using the following:

\( x = \frac{u + v}{2}, \quad y = \frac{u - v}{2} \)

Step 2: Form the Jacobian matrix

The Jacobian matrix is given by:

\[ J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} \]

Substituting the values:

\[ J = \begin{bmatrix} \frac{\partial}{\partial u}\left(\frac{u+v}{2}\right) & \frac{\partial}{\partial v}\left(\frac{u+v}{2}\right) \\ \frac{\partial}{\partial u}\left(\frac{u-v}{2}\right) & \frac{\partial}{\partial v}\left(\frac{u-v}{2}\right) \end{bmatrix} \]

\[ J = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \]

Step 3: Calculate the Jacobian Determinant

The Jacobian determinant is given by:

\[ \det(J) = \left(\frac{1}{2} \cdot -\frac{1}{2}\right) - \left(\frac{1}{2} \cdot \frac{1}{2}\right) \]

\[ \det(J) = -\frac{1}{4} - \frac{1}{4} = -\frac{2}{4} = -\frac{1}{2} \]

✅ Final Answer:

The correct answer is (b) -1/2.

5. The area of the region bounded by \(y = x^2\) and \(y = 2x\) is given by:

Correct Answer: a) \(\int_{0}^{2} \int_{x^2}^{2x} \,dy\,dx\)

Explanation:

To find the area of the region bounded by the curves \( y = x^2 \) and \( y = 2x \), follow these steps:

  1. First, find their points of intersection by solving the equation:
    2. \[ x^2 = 2x \] Rearrange it: \[ x^2 - 2x = 0 \quad \Longrightarrow \quad x(x - 2) = 0 \] The points of intersection are \( x = 0 \) and \( x = 2 \).
  2. Next, set up the double integral. The lower curve is \( y = x^2 \) and the upper curve is \( y = 2x \).
  3. For a vertical slice (integrating with respect to \( y \) first), the bounds for \( y \) are from the lower curve to the upper curve:
    4. \[ x^2 \leq y \leq 2x \]
  4. The bounds for \( x \) are from 0 to 2, as calculated earlier. Hence, the correct integral is:
    5. \[ \int_{0}^{2} \int_{x^2}^{2x} \, dy \, dx \] ✅ The correct answer is **(a)**.

6. The volume under the surface z = 4 - x² - y² and above the xy-plane is:

Correct Answer: b) 8π

Explanation:

To find the volume under the surface \( z = 4 - x^2 - y^2 \) and above the xy-plane, follow these steps:

  1. Recognize that the equation \( z = 4 - x^2 - y^2 \) represents an inverted paraboloid.
  2. Convert to polar coordinates using:
    3. \[ x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta \]
  3. In polar form, the surface becomes:
    4. \[ z = 4 - r^2 \]
  4. Find the boundary by setting \( z = 0 \) to get the radius:
    5. \[ 4 - r^2 = 0 \quad \Longrightarrow \quad r = 2 \]
  5. Set up the volume integral:
    6. \[ V = \int_0^{2\pi} \int_0^2 (4 - r^2) \cdot r \, dr \, d\theta \]
  6. Evaluate the inner integral:
    7. \[ \int_0^2 (4r - r^3) \, dr = \left[ 2r^2 - \frac{r^4}{4} \right]_0^2 = 8 - 4 = 4 \]
  7. Now multiply by the outer integral:
    8. \[ V = \int_0^{2\pi} 4 \, d\theta = 4 \cdot 2\pi = 8\pi \] The correct answer is b

7. For a lamina with density ρ(x,y) = xy bounded by x=0, y=0, and x+y=1, the mass is:

Correct Answer: a) 1/24

Explanation:

The mass of the lamina is calculated using the double integral:

\[ M = \int \int_D \rho(x, y) \, dA \]
  1. The region \( D \) is bounded by:
    • \( x = 0 \)
    • \( y = 0 \)
    • \( x + y = 1 \)
  2. Expressing the bounds: - \( x \) ranges from 0 to 1. - \( y \) ranges from 0 to \( 1 - x \).
  3. Set up the double integral:
    4. \[ M = \int_0^1 \int_0^{1-x} xy \, dy \, dx \]
  4. Evaluate the inner integral with respect to \( y \):
    5. \[ \int_0^{1-x} xy \, dy = x \left[ \frac{y^2}{2} \right]_0^{1-x} = x \cdot \frac{(1-x)^2}{2} \]
  5. Now evaluate the outer integral:
    6. \[ M = \int_0^1 \frac{x (1-x)^2}{2} \, dx \]
  6. Expand and integrate:
    7. \[ M = \frac{1}{2} \int_0^1 x (1 - 2x + x^2) \, dx \] \[ M = \frac{1}{2} \left[ \int_0^1 x \, dx - 2 \int_0^1 x^2 \, dx + \int_0^1 x^3 \, dx \right] \]
  7. Using standard integrals:
    8. \[ \int_0^1 x \, dx = \frac{1}{2}, \quad \int_0^1 x^2 \, dx = \frac{1}{3}, \quad \int_0^1 x^3 \, dx = \frac{1}{4} \]
  8. Substituting:
    9. \[ M = \frac{1}{2} \left( \frac{1}{2} - 2 \cdot \frac{1}{3} + \frac{1}{4} \right) = \frac{1}{2} \left( \frac{6}{12} - \frac{8}{12} + \frac{3}{12} \right) \] \[ M = \frac{1}{2} \cdot \frac{1}{12} = \frac{1}{12} \] ✅ The correct answer is **(b)**.

8. The x-coordinate of the center of mass of the region bounded by y=x² and y=4 with constant density is:

Correct Answer: a) 0

Explanation:

We are tasked with finding the x-coordinate of the center of mass of the region bounded by:

  • y = x²
  • y = 4
  • Constant density

Step 1: Formula for the x-coordinate of the Center of Mass

The formula is:

\[ \bar{x} = \frac{\int\int_D x \, dA}{\int\int_D \, dA} \]

Step 2: Determine the Bounds of Integration

From \( y = x^2 \rightarrow x = \sqrt{y} \) and \( y = 4 \), we find:

  • \( x = -\sqrt{y} \text{ to } x = \sqrt{y} \)
  • \( y = 0 \text{ to } y = 4 \)

Step 3: Calculate the Area (Denominator)

\[ A = \int_0^4 \int_{-\sqrt{y}}^{\sqrt{y}} 1 \, dx \, dy \]

First, integrate with respect to \(x\):

\[ \int_{-\sqrt{y}}^{\sqrt{y}} 1 \, dx = 2\sqrt{y} \]

Then integrate with respect to \(y\):

\[ A = \int_0^4 2\sqrt{y} \, dy \]

We know:

\[ \int y^{1/2} \, dy = \frac{2}{3}y^{3/2} \]

\[ A = 2 \cdot \frac{2}{3} \cdot 4^{3/2} = 2 \cdot \frac{2}{3} \cdot 8 = \frac{32}{3} \]

Step 4: Calculate the First Moment (Numerator)

\[ \int_0^4 \int_{-\sqrt{y}}^{\sqrt{y}} x \, dx \, dy \]

Since the function \(x\) is odd and the region is symmetric about the y-axis:

\[ \int_{-\sqrt{y}}^{\sqrt{y}} x \, dx = 0 \]

Step 5: Calculate the x-Coordinate of the Center of Mass

\[ \bar{x} = \frac{0}{\frac{32}{3}} = 0 \]

✅ Final Answer:

The correct choice is (a) 0.

9. For the transformation \( u = xy \), \( v = \frac{y}{x} \), the Jacobian \( \frac{\partial(x,y)}{\partial(u,v)} \) is:

Correct Answer: a) \( \frac{1}{2v} \)

Explanation:

We are tasked with finding the Jacobian determinant of the transformation:

\[ u = xy, \quad v = \frac{y}{x} \]

First, we express \(x\) and \(y\) in terms of \(u\) and \(v\):

\[ x = \sqrt{\frac{u}{v}}, \quad y = v \sqrt{\frac{u}{v}} \]

Step 1: Form the Jacobian Matrix

\[ J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} \]

Step 2: Compute the Partial Derivatives

\[ \frac{\partial x}{\partial u} = \frac{1}{2\sqrt{uv}}, \quad \frac{\partial x}{\partial v} = -\frac{\sqrt{u}}{2v^{3/2}} \] \[ \frac{\partial y}{\partial u} = \frac{v}{2\sqrt{uv}}, \quad \frac{\partial y}{\partial v} = \sqrt{\frac{u}{v}} - \frac{u}{2v^{3/2}} \]

Step 3: Calculate the Jacobian Determinant

\[ \left| J \right| = \frac{1}{2\sqrt{uv}} \left(\sqrt{\frac{u}{v}} - \frac{u}{2v^{3/2}}\right) - \left(-\frac{\sqrt{u}}{2v^{3/2}}\right) \cdot \frac{v}{2\sqrt{uv}} \] \[ \left| J \right| = \frac{1}{2v} \] ✅ The correct answer is **(a)**.

10. The volume of the solid bounded by \( z = \sqrt{x^2+y^2} \) and \( z = 2 \) is:

Correct Answer: b) 8π/3

Explanation:

The given solid is defined by:

\[ z = \sqrt{x^2+y^2}, \quad z = 2 \] This describes a cone with its vertex at the origin and a height of 2. The cone's base is at \(z = 2\) with radius 2.

Step 1: Use the Volume Formula for a Cone

The volume of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] Here, the radius \(r = 2\) and the height \(h = 2\). \[ V = \frac{1}{3} \pi (2)^2 (2) = \frac{1}{3} \pi \cdot 4 \cdot 2 = \frac{8\pi}{3} \] ✅ The correct answer is **(b)**.