MT2176 Further Calculus

Correct Answer: (b)

Explanation: 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 computed as an iterated integral, and the order of integration can be switched.
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Chapter 5, Theorem 5.1, p. 147.

Correct Answer: (b)

Explanation: We first integrate with respect to \(x\): \(\int_0^2 (x^2 + y) dx = [\frac{x^3}{3} + yx]_0^2 = (\frac{8}{3} + 2y) - 0 = \frac{8}{3} + 2y\). Then, we integrate the result with respect to \(y\): \(\int_0^1 (\frac{8}{3} + 2y) dy = [\frac{8}{3}y + y^2]_0^1 = (\frac{8}{3} + 1) - 0 = \frac{11}{3}\).
Source: Based on methods for finding volumes over rectangular bases. See Ostaszewski (1991), Chapter 5, Example 5.2, p. 147 for a similar problem.

Correct Answer: (b)

Explanation: The volume is given by the double integral \(\iint_R (2x+y) dA\). We can compute this as an iterated integral: \(V = \int_0^3 \int_0^2 (2x+y) dx dy\). Inner integral: \(\int_0^2 (2x+y) dx = [x^2 + yx]_0^2 = 4 + 2y\). Outer integral: \(\int_0^3 (4+2y) dy = [4y + y^2]_0^3 = (12 + 9) - 0 = 21\).
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9, on Double Integrals.

Correct Answer: (d)

Explanation: The region of integration is defined by \(x \le y \le x^2\) for \(0 \le x \le 1\). However, for \(x\) in the interval \((0, 1)\), we have \(x^2 < x\). This means there are no \(y\) values that satisfy \(x \le y \le x^2\). The region of integration only consists of the points (0,0) and (1,1). Therefore, the integral over this region is 0.
Source: Based on the principles of setting up iterated integrals over general regions. See Ostaszewski (1991), Chapter 5, Section 5.1.5.

Correct Answer: (a)

Explanation: The region of integration is \(y \le x \le 1\) and \(0 \le y \le 1\). This is a triangle with vertices (0,0), (1,0), and (1,1). Changing the order of integration, the limits become \(0 \le y \le x\) and \(0 \le x \le 1\). The integral becomes: \(\int_0^1 \int_0^x e^{x^2} dy dx = \int_0^1 [y e^{x^2}]_0^x dx = \int_0^1 x e^{x^2} dx\). Using substitution \(u = x^2\), \(du = 2x dx\): \(\frac{1}{2} \int_0^1 e^u du = \frac{1}{2} [e^u]_0^1 = \frac{1}{2}(e - 1)\).
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Example 5.8, p. 156.

Correct Answer: (b)

Explanation: The transformation is \(x = r \cos \theta\) 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: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Example 5.14, p. 157.

Correct Answer: (a)

Explanation: In polar coordinates, \(x^2+y^2 = r^2\) and \(dA = r dr d\theta\). The region \(D\) is described by \(0 \le r \le R\) and \(0 \le \theta \le \frac{\pi}{2}\). The integral becomes: \[ \int_0^{\pi/2} \int_0^R e^{-r^2} r dr d\theta \] Let \(u = -r^2\), so \(du = -2r dr\). The inner integral is: \[ \int_0^{-R^2} e^u (-\frac{1}{2} du) = -\frac{1}{2} [e^u]_0^{-R^2} = -\frac{1}{2}(e^{-R^2} - 1) = \frac{1}{2}(1 - e^{-R^2}) \] The outer integral is: \[ \int_0^{\pi/2} \frac{1}{2}(1 - e^{-R^2}) d\theta = \frac{1}{2}(1 - e^{-R^2}) [\theta]_0^{\pi/2} = \frac{\pi}{4}(1 - e^{-R^2}) \]
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Example 5.14, p. 157.

Correct Answer: (a)

Explanation: The volume is given by \(\int_0^1 \int_0^x x dy dx\). Inner integral: \(\int_0^x x dy = [xy]_0^x = x^2\). Outer integral: \(\int_0^1 x^2 dx = [\frac{x^3}{3}]_0^1 = \frac{1}{3}\).
Source: Based on principles from Ostaszewski (1991), Chapter 5, Section 5.1.5.

Correct Answer: (b)

Explanation: The region of integration is defined by \(0 \le x \le 3\) and \(0 \le y \le 2\) (since \(z=0 \implies 4-y^2=0 \implies y=2\)). The volume is \(V = \int_0^3 \int_0^2 (4-y^2) dy dx\). The inner integral is \(\int_0^2 (4-y^2) dy = [4y - \frac{y^3}{3}]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}\). The outer integral is \(\int_0^3 \frac{16}{3} dx = \frac{16}{3} [x]_0^3 = 16\).
Source: Based on principles from Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9.

Correct Answer: (a)

Explanation: The region is bounded by \(y=x^2\) and \(y=2x\). The intersection points are \(x^2=2x \implies x(x-2)=0\), so \(x=0, x=2\). The region is \(x^2 \le y \le 2x\) for \(0 \le x \le 2\). To reverse the order, we express \(x\) in terms of \(y\). From \(y=2x\), we get \(x=y/2\). From \(y=x^2\), we get \(x=\sqrt{y}\). The range for \(y\) is from 0 to 4. For a fixed \(y\) between 0 and 4, \(x\) goes from \(y/2\) to \(\sqrt{y}\). So the integral is \(\int_0^4 \int_{y/2}^{\sqrt{y}} f(x,y) dx dy\).
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Example 5.8, p. 156.

Correct Answer: (a)

Explanation: We compute the partial derivatives: \(\frac{\partial u}{\partial x} = 1\), \(\frac{\partial u}{\partial y} = 1\), \(\frac{\partial v}{\partial x} = -\frac{y}{x^2}\), \(\frac{\partial v}{\partial y} = \frac{1}{x}\). The Jacobian is: \[ \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} 1 & 1 \\ -y/x^2 & 1/x \end{vmatrix} = (1)(\frac{1}{x}) - (1)(-\frac{y}{x^2}) = \frac{1}{x} + \frac{y}{x^2} = \frac{x+y}{x^2} \]
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Example 5.12, p. 155.

Correct Answer: (b)

Explanation: The region is bounded by \(y=x^2\) and \(y=x\) for \(0 \le x \le 1\). The integral is: \[ \int_0^1 \int_{x^2}^{x} x dy dx = \int_0^1 [xy]_{x^2}^{x} dx = \int_0^1 (x^2 - x^3) dx \] \[ = [\frac{x^3}{3} - \frac{x^4}{4}]_0^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \]
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9, Solved Problems.

Correct Answer: (a)

Explanation: We compute the partial derivatives: \(\frac{\partial u}{\partial x} = 2x\), \(\frac{\partial u}{\partial y} = -2y\), \(\frac{\partial v}{\partial x} = 2y\), \(\frac{\partial v}{\partial y} = 2x\). The Jacobian is: \[ \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} 2x & -2y \\ 2y & 2x \end{vmatrix} = (2x)(2x) - (-2y)(2y) = 4x^2 + 4y^2 = 4(x^2+y^2) \]
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9, Solved Problems.

Correct Answer: (d)

Explanation: The volume is under the surface \(z = c(1 - x/a - y/b)\). The region of integration in the xy-plane is the triangle bounded by \(x=0, y=0\) and \(x/a + y/b = 1\). This can be written as \(0 \le x \le a\) and \(0 \le y \le b(1-x/a)\). The volume integral is therefore \(\iint_D c(1 - x/a - y/b) dA\), which becomes the iterated integral \(c \int_0^a \int_0^{b(1-x/a)} (1 - x/a - y/b) dy dx\).
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9.

Correct Answer: (a)

Explanation: \(\int_0^1 \int_0^2 (x^2+2xy+y^2) dx dy = \int_0^1 [\frac{x^3}{3} + x^2y + xy^2]_0^2 dy = \int_0^1 (\frac{8}{3} + 4y + 2y^2) dy = [\frac{8y}{3} + 2y^2 + \frac{2y^3}{3}]_0^1 = \frac{8}{3} + 2 + \frac{2}{3} = \frac{10}{3} + 2 = \frac{16}{3}\).
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9.

Correct Answer: (c)

Explanation: The region is bounded by \(y=x^2\) and \(y=2x\) for \(0 \le x \le 2\). The intersection points are (0,0) and (2,4). To reverse the order, we express x in terms of y. The curves are \(x = \sqrt{y}\) and \(x = y/2\). The y-range is from 0 to 4. For a fixed y, x ranges from the line \(x=y/2\) to the parabola \(x=\sqrt{y}\). Therefore, the integral is \(\int_0^4 \int_{y/2}^{\sqrt{y}} dx dy\).
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Chapter 5, Section 5.1.5.

Correct Answer: (a)

Explanation: We compute the partial derivatives: \(\frac{\partial x}{\partial u} = \cosh v\), \(\frac{\partial x}{\partial v} = u \sinh v\), \(\frac{\partial y}{\partial u} = \sinh v\), \(\frac{\partial y}{\partial v} = u \cosh v\). The Jacobian is: \[ \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \cosh v & u \sinh v \\ \sinh v & u \cosh v \end{vmatrix} = u \cosh^2 v - u \sinh^2 v = u(\cosh^2 v - \sinh^2 v) = u(1) = u \]
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9.

Correct Answer: (b)

Explanation: Using polar coordinates, the region is \(2 \le r \le 3\) and \(0 \le \theta \le \pi\). The integral is \(\int_0^\pi \int_2^3 (r \sin\theta) r dr d\theta = (\int_0^\pi \sin\theta d\theta) (\int_2^3 r^2 dr) = [-\cos\theta]_0^\pi [\frac{r^3}{3}]_2^3 = (2) (\frac{19}{3}) = \frac{38}{3}\).
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9.

Correct Answer: (c)

Explanation: The sphere is \(x^2+y^2+z^2=a^2\). The volume is twice the volume of the upper hemisphere, where \(z = \sqrt{a^2-x^2-y^2}\). In polar coordinates, this is \(z = \sqrt{a^2-r^2}\). The volume is \(V = \iint_D 2\sqrt{a^2-x^2-y^2} dA\) where D is the disk \(x^2+y^2 \le a^2\). In polar coordinates, this becomes \(2 \int_0^{2\pi} \int_0^a \sqrt{a^2-r^2} r dr d\theta\).
Source: Binmore, K., & Davies, J. (2002). Calculus: concepts and methods. Cambridge University Press. Chapter 11.

Correct Answer: (a)

Explanation: This is the integral over the first quadrant. In polar coordinates, it becomes \(\int_0^{\pi/2} \int_0^\infty e^{-r^2} r dr d\theta\). The inner integral is \(\int_0^\infty e^{-r^2} r dr = [-\frac{1}{2}e^{-r^2}]_0^\infty = 0 - (-\frac{1}{2}) = \frac{1}{2}\). The outer integral is \(\int_0^{\pi/2} \frac{1}{2} d\theta = \frac{1}{2} [\theta]_0^{\pi/2} = \frac{\pi}{4}\).
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Example 5.15, p. 163.

Correct Answer: (d)

Explanation: Transform the vertices: (1,0) -> (u=1, v=1). (0,1) -> (u=-1, v=1). (-1,0) -> (u=-1, v=-1). (0,-1) -> (u=1, v=-1). The transformed region is a square with these vertices.
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9.

Correct Answer: (b)

Explanation: The inverse transformation is \(x = (u+v)/2\), \(y = (u-v)/2\). The Jacobian is \(\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{vmatrix} = -1/4 - 1/4 = -1/2\).
Source: Ostaszewski, A. (1991). Advanced mathematical methods. Cambridge University Press. Chapter 5, Section 5.2.

Correct Answer: (c)

Explanation: The integral of \(x+y\) is 1/3. So the integral of \(2(x+y)\) is 2/3.
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9.

Correct Answer: (a)

Explanation: Find intersection points: \(x^2 = x+2 \implies x^2-x-2=0 \implies (x-2)(x+1)=0\). So \(x=-1, x=2\). The area is \(\int_{-1}^2 (x+2 - x^2) dx = [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^2 = (2+4-\frac{8}{3}) - (\frac{1}{2}-2+\frac{1}{3}) = (6-\frac{8}{3}) - (-\frac{7}{6}) = \frac{10}{3} + \frac{7}{6} = \frac{20+7}{6} = \frac{27}{6} = \frac{9}{2}\).
Source: Binmore, K., & Davies, J. (2002). Calculus: concepts and methods. Cambridge University Press. Chapter 11.

Correct Answer: (c)

Explanation: The Jacobian determinant is given by: \[ \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} \] \[ = (2)(2) - (-3)(1) = 4 + 3 = 7 \]
Source: Wrede, R. C., & Spiegel, M. (2010). Schaum's outline of advanced calculus. McGraw-Hill. Chapter 9, Transformations of Multiple Integrals.