MT2176 Further Calculus: Double Integrals Quiz

Correct Answer: B

Fubini’s Theorem states that if \(f(x, y)\) is continuous on a rectangle \(R = [a, b] \times [c, d]\), then the double integral can be evaluated as a repeated integral. The order of integration does not matter.

$$ \iint_R f(x, y) \,dA = \int_c^d \left[ \int_a^b f(x, y) \,dx \right] \,dy = \int_a^b \left[ \int_c^d f(x, y) \,dy \right] \,dx $$

Source: MT2176 Subject Guide, Chapter 5, Section 5.1.4, p. 141.

Correct Answer: C

We compute the iterated integral:

$$ \int_0^2 \int_0^1 (x^2 + y^2) \,dx \,dy = \int_0^2 \left[ \frac{x^3}{3} + y^2x \right]_0^1 \,dy $$

$$ = \int_0^2 \left( \frac{1}{3} + y^2 \right) \,dy = \left[ \frac{1}{3}y + \frac{y^3}{3} \right]_0^2 = \frac{2}{3} + \frac{8}{3} = \frac{10}{3} $$

Source: This is a standard application of calculating volumes over rectangular bases. See MT2176 Subject Guide, Section 5.1.1, p. 136.

Correct Answer: D

The volume is given by the double integral. Since the integrand does not depend on x, the calculation is straightforward:

$$ V = \int_0^1 \int_0^1 e^{-y} \,dx \,dy = \int_0^1 \left[ xe^{-y} \right]_0^1 \,dy = \int_0^1 e^{-y} \,dy $$

$$ = \left[ -e^{-y} \right]_0^1 = -e^{-1} - (-e^0) = 1 - e^{-1} $$

Source: Inspired by Example 5.1 and 5.2 from the MT2176 Subject Guide, p. 141-142.

Correct Answer: A

For a general region bounded by curves as functions of x, the inner integral must be with respect to y, with the bounding curves as its limits. The outer integral is then with respect to x, with the constant bounds.

Source: MT2176 Subject Guide, Section 5.1.5, p. 143-152.

Correct Answer: C

First, integrate with respect to y:

$$ \int_0^x (x+y) \,dy = \left[ xy + \frac{y^2}{2} \right]_0^x = x(x) + \frac{x^2}{2} = x^2 + \frac{x^2}{2} = \frac{3}{2}x^2 $$

Then, integrate the result with respect to x:

$$ \int_0^1 \frac{3}{2}x^2 \,dx = \frac{3}{2} \left[ \frac{x^3}{3} \right]_0^1 = \frac{3}{2} \left( \frac{1}{3} \right) = \frac{1}{2} $$

Wait, let me recheck the calculation. Inner integral: \(xy + y^2/2\) from 0 to x is \(x^2 + x^2/2 = 3x^2/2\). Outer integral: \(\int_0^1 (3/2)x^2 dx = (3/2) [x^3/3]_0^1 = (3/2)(1/3) = 1/2\). Let me re-evaluate. Ah, I see the mistake in my scratchpad. The answer should be 1/2. Let me re-evaluate the options. It seems I made a mistake in the options provided in my thought process. Let me correct the calculation. \(\int_0^1 [xy + y^2/2]_0^x dx = \int_0^1 (x^2 + x^2/2) dx = \int_0^1 (3/2)x^2 dx = [x^3/2]_0^1 = 1/2\). Let me check the provided solution. The solution states 2/3. Let me re-read the question. Ah, I see my scratchpad calculation was correct, but I mis-typed the final answer. Let me re-calculate one more time. Inner integral: \(\int_0^x (x+y) dy = [xy + \frac{y^2}{2}]_0^x = x^2 + \frac{x^2}{2} = \frac{3x^2}{2}\). Outer integral: \(\int_0^1 \frac{3x^2}{2} dx = \frac{3}{2} [\frac{x^3}{3}]_0^1 = \frac{3}{2} \cdot \frac{1}{3} = \frac{1}{2}\). The correct answer is 1/2. I will mark 'b' as correct and regenerate the explanation.

Correct Answer: B

First, integrate with respect to y:

$$ \int_0^x (x+y) \,dy = \left[ xy + \frac{y^2}{2} \right]_0^x = x(x) + \frac{x^2}{2} - 0 = \frac{3}{2}x^2 $$

Then, integrate the result with respect to x:

$$ \int_0^1 \frac{3}{2}x^2 \,dx = \frac{3}{2} \left[ \frac{x^3}{3} \right]_0^1 = \frac{3}{2} \left( \frac{1}{3} - 0 \right) = \frac{1}{2} $$

Source: Based on concepts from Ostaszewski, Advanced Mathematical Methods, Chapter 19.

Correct Answer: B

The transformation is given by \(x = r \cos \theta\) and \(y = r \sin \theta\). The Jacobian determinant is:

$$ \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos \theta & -r \sin \theta \ \sin \theta & r \cos \theta \end{vmatrix} $$

$$ = (\cos \theta)(r \cos \theta) - (-r \sin \theta)(\sin \theta) = r \cos^2 \theta + r \sin^2 \theta = r(\cos^2 \theta + \sin^2 \theta) = r $$

Source: Wrede & Spiegel, Schaum\'s Outlines: Advanced Calculus, Chapter 9.

Correct Answer: D

The volume is given by the double integral. Since the integrand does not depend on x, the calculation is straightforward:

$$ V = \int_0^1 \int_0^1 e^{-y} \,dx \,dy = \int_0^1 \left[ xe^{-y} \right]_0^1 \,dy = \int_0^1 e^{-y} \,dy $$

$$ = \left[ -e^{-y} \right]_0^1 = -e^{-1} - (-e^0) = 1 - e^{-1} $$

Source: Inspired by Example 5.1 and 5.2 from the MT2176 Subject Guide, p. 141-142.

Correct Answer: A

For a general region bounded by curves as functions of x, the inner integral must be with respect to y, with the bounding curves as its limits. The outer integral is then with respect to x, with the constant bounds.

Source: MT2176 Subject Guide, Section 5.1.5, p. 143-152.

Correct Answer: B

First, integrate with respect to y:

$$ \int_0^x (x+y) \,dy = \left[ xy + \frac{y^2}{2} \right]_0^x = x(x) + \frac{x^2}{2} - 0 = \frac{3}{2}x^2 $$

Then, integrate the result with respect to x:

$$ \int_0^1 \frac{3}{2}x^2 \,dx = \frac{3}{2} \left[ \frac{x^3}{3} \right]_0^1 = \frac{3}{2} \left( \frac{1}{3} - 0 \right) = \frac{1}{2} $$

Source: Based on concepts from Ostaszewski, Advanced Mathematical Methods, Chapter 19.

Correct Answer: B

The transformation is given by \(x = r \cos \theta\) and \(y = r \sin \theta\). The Jacobian determinant is:

$ \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos \theta & -r \sin \theta \ \sin \theta & r \cos \theta \end{vmatrix} $

$ = (\cos \theta)(r \cos \theta) - (-r \sin \theta)(\sin \theta) = r \cos^2 \theta + r \sin^2 \theta = r(\cos^2 \theta + \sin^2 \theta) = r $

Source: Wrede & Spiegel, Schaum\'s Outlines: Advanced Calculus, Chapter 9.

Correct Answer: A

The limits of integration tell us how the region is defined. The outer integral \(\int_0^2 ... dx\) means x varies from 0 to 2. The inner integral \(\int_0^{x^2} ... dy\) means for each x, y varies from 0 to \(x^2\). This describes the area under the parabola \(y=x^2\) from \(x=0\) to \(x=2\), bounded below by the x-axis (y=0).

Source: MT2176 Subject Guide, Section 5.1.5, p. 143.

Correct Answer: C

First, evaluate the inner integral with respect to y:

$ \int_0^{\sin x} y \,dy = \left[ \frac{y^2}{2} \right]_0^{\sin x} = \frac{\sin^2 x}{2} $

Now, evaluate the outer integral. Using the identity \(\sin^2 x = \frac{1 - \cos(2x)}{2}\):

$ \int_0^\pi \frac{\sin^2 x}{2} \,dx = \frac{1}{4} \int_0^\pi (1 - \cos(2x)) \,dx = \frac{1}{4} \left[ x - \frac{\sin(2x)}{2} \right]_0^\pi $

$ = \frac{1}{4} \left( (\pi - \frac{\sin(2\pi)}{2}) - (0 - \frac{\sin(0)}{2}) \right) = \frac{1}{4}(\pi - 0 - 0) = \frac{\pi}{4} $

Source: Based on concepts from Wrede & Spiegel, Schaum\'s Outlines: Advanced Calculus, Chapter 9.

Correct Answer: C

The region is given by \(0 \le y \le \sqrt{x}\) and \(0 \le x \le 4\). This means \(y^2 \le x\). The y-values range from \(y=0\) to \(y=\sqrt{4}=2\). For a fixed y in [0, 2], x ranges from \(y^2\) to 4. So the integral becomes \(\int_0^2 \int_{y^2}^4 f(x, y) \,dx \,dy\).

Source: MT2176 Subject Guide, Section 5.1.5, p. 151.

Correct Answer: A

First, we find the Jacobian of the given transformation, \(\frac{\partial(u, v)}{\partial(x, y)}\).

$ \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} 1 & 1 \ 1 & -1 \end{vmatrix} = -1 - 1 = -2 $

The required Jacobian is the inverse of this:

$ \frac{\partial(x, y)}{\partial(u, v)} = \left( \frac{\partial(u, v)}{\partial(x, y)} \right)^{-1} = \frac{1}{-2} = -\frac{1}{2} $

Source: Ostaszewski, Advanced Mathematical Methods, Chapter 19, Section 19.6.

Correct Answer: B

A double integral of a function \(f(x,y)\) over a region R gives the volume of the solid under the surface \(z=1-x^2-y^2\) and above the region R in the xy-plane. Here, the function is \(z=1-x^2-y^2\) (an inverted paraboloid) and the region is the unit square \([0,1] \times [0,1]\).

Source: MT2176 Subject Guide, Section 5.1.1, p. 136.

Correct Answer: C

We can separate the integrals since the limits are constant and the function is separable:

$ \left( \int_1^2 x^2 dx \right) \left( \int_0^1 y dy \right) = \left[ \frac{x^3}{3} \right]_1^2 \left[ \frac{y^2}{2} \right]_0^1 $

$ = \left( \frac{8}{3} - \frac{1}{3} \right) \left( \frac{1}{2} - 0 \right) = \left( \frac{7}{3} \right) \left( \frac{1}{2} \right) = \frac{7}{6} $

Source: Wrede & Spiegel, Schaum\'s Outlines: Advanced Calculus, Chapter 9.

Correct Answer: D

The volume is \(\iint_D (x^2+y^2) dA\), where D is the disk \(x^2+y^2 \le 4\). It is best to use polar coordinates. The transformation is \(x=r\cos\theta, y=r\sin\theta\), so \(x^2+y^2=r^2\). The Jacobian is \(r\). The region D is \(0 \le r \le 2\) and \(0 \le \theta \le 2\pi\). The integral becomes:

$ \int_0^{2\pi} \int_0^2 (r^2) \cdot r \,dr \,d\theta = \int_0^{2\pi} \int_0^2 r^3 \,dr \,d\theta $

Source: MT2176 Subject Guide, Section 5.2, p. 152.

Correct Answer: A

We evaluate the iterated integral:

$ \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_0^2 \,d\theta = \int_0^{2\pi} \frac{16}{4} \,d\theta = \int_0^{2\pi} 4 \,d\theta $

$ = \left[ 4\theta \right]_0^{2\pi} = 4(2\pi) = 8\pi $

Source: Wrede & Spiegel, Schaum\'s Outlines: Advanced Calculus, Chapter 9.

Correct Answer: C

The region of integration is defined by \(0 \le x \le \sqrt{1-y^2}\) and \(0 \le y \le 1\). This is the quarter-circle in the first quadrant with radius 1. Squaring the first inequality gives \(x^2 \le 1-y^2 \Rightarrow x^2+y^2 \le 1\). To reverse the order, we let x vary from 0 to 1. For a fixed x, y varies from 0 to \(\sqrt{1-x^2}\). The integral becomes \(\int_0^1\int_0^{\sqrt{1-x^2}} f(x,y) dy dx\).

Source: Ostaszewski, Advanced Mathematical Methods, Chapter 19.

Correct Answer: B

We compute the Jacobian determinant directly from the given transformation.

$$ \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1/v & -u/v^2 \ 0 & 1 \end{vmatrix} $$

$$ = (1/v)(1) - (-u/v^2)(0) = 1/v $$

Source: Based on concepts from MT2176 Subject Guide, Section 5.2, p. 152.

Correct Answer: A

The limits of integration tell us how the region is defined. The outer integral \(\int_0^2 ... dx\) means x varies from 0 to 2. The inner integral \(\int_0^{x^2} ... dy\) means for each x, y varies from 0 to \(x^2\). This describes the area under the parabola \(y=x^2\) from \(x=0\) to \(x=2\), bounded below by the x-axis (y=0).

Source: MT2176 Subject Guide, Section 5.1.5, p. 143.

Correct Answer: D

First, evaluate the inner integral with respect to y:

$$ \int_0^{\sin x} y \,dy = \left[ \frac{y^2}{2} \right]_0^{\sin x} = \frac{\sin^2 x}{2} $$

Now, evaluate the outer integral:

$$ \int_0^\pi \frac{\sin^2 x}{2} \,dx = \frac{1}{2} \int_0^\pi \frac{1 - \cos(2x)}{2} \,dx = \frac{1}{4} \left[ x - \frac{\sin(2x)}{2} \right]_0^\pi $$

$$ = \frac{1}{4} \left( (\pi - 0) - (0 - 0) \right) = \frac{\pi}{4} $$

Let me re-check. Ah, the final calculation is \(\frac{1}{4} [\pi] = \frac{\pi}{4}\). Let me re-read the question and my calculation. \(\int_0^\pi \frac{\sin^2 x}{2} dx = \frac{1}{4} \int_0^\pi (1 - \cos(2x)) dx = \frac{1}{4} [x - \frac{1}{2}\sin(2x)]_0^\pi = \frac{1}{4} ((\pi - 0) - (0-0)) = \frac{\pi}{4}\). The answer is \(\pi/4\). Let me check the options again. It seems I have made a mistake in the provided answer key in my thought process. The correct answer is C. Let me regenerate the explanation with the correct answer.

Correct Answer: C

First, evaluate the inner integral with respect to y:

$$ \int_0^{\sin x} y \,dy = \left[ \frac{y^2}{2} \right]_0^{\sin x} = \frac{\sin^2 x}{2} $$

Now, evaluate the outer integral. Using the identity \(\sin^2 x = \frac{1 - \cos(2x)}{2}\):

$$ \int_0^\pi \frac{\sin^2 x}{2} \,dx = \frac{1}{4} \int_0^\pi (1 - \cos(2x)) \,dx = \frac{1}{4} \left[ x - \frac{\sin(2x)}{2} \right]_0^\pi $$

$$ = \frac{1}{4} \left( (\pi - \frac{\sin(2\pi)}{2}) - (0 - \frac{\sin(0)}{2}) \right) = \frac{1}{4}(\pi - 0 - 0) = \frac{\pi}{4} $$

Source: Based on concepts from Wrede & Spiegel, Schaum\'s Outlines: Advanced Calculus, Chapter 9.

Correct Answer: A

The region of integration is defined by \(x \le y \le 1\) and \(0 \le x \le 1\). This is a triangle with vertices at (0,0), (1,1), and (0,1). To change the order of integration, we fix y first. The y-values range from 0 to 1. For a fixed y, x ranges from 0 to y. Therefore, the new integral is:

$$ \int_0^1 \int_0^y f(x, y) \,dx \,dy $$

Source: This is a standard problem on changing the order of integration. See Wrede & Spiegel, Schaum's Outlines: Advanced Calculus, Chapter 9.