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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 6.5 Week 6 Activity

Subsection 6.5.1 Topology

Activity 6.5.1.

Consider the set \([0,5] \times [0,5] \times [0,5]\) in \(\RR^3\text{.}\) For each of the following points, determine if the point is a boundary point, an interior point, or neither.
  1. \(\displaystyle (3,3,3)\)
  2. \(\displaystyle (0,0,0)\)
  3. \(\displaystyle (0,5,0)\)
  4. \(\displaystyle (5,4,-1)\)
  5. \(\displaystyle (3,2,5)\)
Is the interval open, closed, or neither?
Solution.
  1. \((3,3,3)\) is an interior point.
  2. \((0,0,0)\) is a boundary point.
  3. \((0,5,0)\) is a bounardy point.
  4. \((5,4,-1)\) is a neither.
  5. \((3,2,5)\) is a boundary point.
The interval is closed, since it is formed from three closed intervals in \(\RR\text{.}\)

Activity 6.5.2.

Consider the set \((0,4) \times (2,4) \times (0,4)\) in \(\RR^3\text{.}\) For each of the following points, determine if the point is a boundary point, an interior point, or neither.
  1. \(\displaystyle (0,0,0)\)
  2. \(\displaystyle (4,4,4)\)
  3. \(\displaystyle (2,1,3)\)
  4. \(\displaystyle (0,2,3)\)
  5. \(\displaystyle (3,3,3)\)
Is the interval open, closed, or neither?
Solution.
  1. \((0,0,0)\) is neither. (It is outside the set, since it is well away from the \(y\) range.)
  2. \((4,4,4)\) is a boundary point.
  3. \((2,1,3)\) is neither (It is outside the set, since it is well away from the \(y\) range.
  4. \((0,2,3)\) is a boundary point.
  5. \((3,3,3)\) is an interior point.
The interval is open, since it is fromed from three open intervals in \(\RR\text{.}\)

Activity 6.5.3.

Consider the set \([1,3] \times (1,4) \times (0,4]\) in \(\RR^3\text{.}\) For each of the following points, determine if the point is a boundary point, an interior point, or neither.
  1. \(\displaystyle (2,2,2)\)
  2. \(\displaystyle (1,1,0)\)
  3. \(\displaystyle (3,2,2)\)
  4. \(\displaystyle (0,0,0)\)
  5. \(\displaystyle (2,1,0)\)
Is the interval open, closed, or neither?
Solution.
  1. \((2,2,2)\) is an interior point.
  2. \((1,1,0)\) is a boundary point.
  3. \((3,2,2)\) is a boundary point.
  4. \((0,0,0)\) is neither.
  5. \((2,1,0)\) is a boundary point.
The interval is neither open or closed. It includes is boundary planes in \(x\text{,}\) but not in \(y\) and only partially in \(z\text{.}\)

Subsection 6.5.2 Iterated Integrals

Activity 6.5.4.

Integrate the function \(f(x,y) = xy^4 \) on the interval \([-1,2] \times [-3,0]\text{.}\)
Solution.
This is an interated integral. I choose to integrate with \(x\) in the inside and \(y\) on the outside. Then the integral proceeds as two consecutive single-variable integrals.
\begin{align*} \int_{-3}^0 \int_{-1}^2 xy^4 dx dy \amp = \int_{-3}^0 \frac{x^2 y^4}{2} \bigg|_{-1}^{2} dy \\ \amp = \int_{-3}^0 \frac{y^4}{2} (2^2 - (-1)^2) dy \\ \amp = \int_{-3}^0 \frac{3y^4}{2} dy = 3 \frac{y^5}{10} \bigg|_{0}^3 = \frac{729}{10} \end{align*}

Activity 6.5.5.

Integrate the function \(f(x,y) = xy \cos (xy^2) \) on the interval \([0,\pi] \times [0,1]\text{.}\)
Solution.
This is an interated integral. I choose to integrate with \(y\) in the inside and \(x\) on the outside. Then the integral proceeds as two consecutive single-variable integrals. The first step uses the substitution \(u = xy^2\) with \(du = 2xydy\text{.}\) (Not that for this substitution, since it is part of the \(y\) integral, \(x\) is treate as a constant. The substitution uses \(u\) to replace the variable \(y\text{.}\))
\begin{align*} \int_{0}^{\pi} \int_0^1 xy \cos (xy^2) dx dy \amp = \int_0^{\pi} \int_{y=0}^{y=1} \frac{1}{2} \cos u du dx \\ \amp = \int_0^{\pi} \frac{- \sin u}{2} \bigg|_{y=0}^{y=1} = \int_0^{\pi} \frac{ -\sin (xy^2)}{2} \bigg|_0^1\\ \amp = \frac{1}{2} \int_0^{\pi} -\sin (x) + (\sin 0) dx = \frac{-1}{2} \int_0^{\pi} \sin (x) dx \\ \amp = \frac{-1}{2} \cos x \bigg|_0^{\pi} = \frac{-1}{2} 2 = -1 \end{align*}

Activity 6.5.6.

Integrate the function \(f(x,y) = xy \ln (xy)\) on the interval \([1,4] \times [1,4]\text{.}\)
Solution.
This is an interated integral. I choose to integrate with \(x\) in the inside and \(y\) on the outside. Then the integral proceeds as two consecutive single-variable integrals. In the first step, I use integration by parts. In two of the three terms in the final step, I also use integration by parts.
\begin{align*} \amp \int_1^4 \int_1^4 xy \ln (xy) \\ \amp = \int_1^4 \left( \frac{x^2y}{2} \ln(xy) \bigg|_1^4 - \int_1^4 \frac{1}{x} \frac{x^2y}{2} dx \right) dy \\ \amp = \int_1^4 \left( 8y \ln (4y) - \frac{y}{2} \ln y - \int_1^4 \frac{xy}{2} dx \right) dy \\ \amp = \int_1^4 \left( 8y\ln (4y) - \frac{y}{2} \ln y - \frac{x^2y}{4} \bigg|_1^4 \right) dy \\ \amp = \int_1^4 8y\ln (4y) - \frac{y}{2} \ln y - \frac{15y}{4} dy \\ \amp = \frac{8y^2}{2} \ln 4y \bigg|_1^4 - \int_1^4 \frac{8y^2}{2} \frac{1}{y} dy - \left( \frac{y^2}{4} \ln y \bigg|_1^4 - \int_1^4 \frac{y^2}{4} \frac{1}{y} dy \right) - \frac{15y^2}{8} \bigg|_1^4\\ \amp = 64 \ln (16) - 4 \ln 4 - \int_1^4 4y dy - 1 \ln 4 + 0 + \int_1^4 \frac{y}{4} dy - \frac{225}{8}\\ \amp = 128 \ln 4 - 4 \ln 4 - 2(15) - \ln 4 + \frac{15}{8} - \frac{225}{8} = 123 \ln 4 - \frac{15}{4} \end{align*}

Activity 6.5.7.

Integrate the function \(f(x,y) = \frac{1}{x^2y^2}\) on the interval \([0,4] \times [1,4]\text{.}\)
Solution.
This is an interated integral. I choose to integrate with \(y\) in the inside and \(x\) on the outside. Then the integral proceeds as two consecutive single-variable integrals. The integral is improper in \(x\text{,}\) so I will have to use limits when I finish the \(x\) integration.
\begin{align*} \int_0^4 \int_1^4 \frac{1}{x^2y^2} dy dx \amp = \int_0^4 \frac{-1}{x^2y} \bigg|_1^4 dx \\ \amp = \int_0^4 \frac{-1}{4x^2} - \frac{-1}{x^2} dx = \int_0^4 \frac{-1 + 4}{4x^2} dx = \int_0^4 \frac{3}{4x^2}\\ \amp = \frac{3}{4} \frac{-1}{x} \bigg|_0^4 = \frac{3}{4} \left( \frac{-1}{4} - \lim_{a \rightarrow 0} \frac{-1}{a} \right) \end{align*}
The limit does not converge, so the integral is not defined.

Activity 6.5.8.

Integrate the function \(f(x,y,z) = 3x^2 + yz^2 + x^3z\) on the interval \([0,2] \times [-3,3] \times [-1,1]\text{.}\)
Solution.
This is a triple interated integral. I choose to integrate with \(x\) in the inside, then \(y\) moving outward, and finally \(z\) on the outside. Then the integral proceeds as two consecutive single-variable integrals.
\begin{align*} \amp \int_{-1}^1 \int_{-3}^3 \int_0^2 3x^2 + 3yz^2 +x^3z dx dy dz \\ \amp = \int_{-1}^1 \int_{-3}^3 \left( x^3 + 3xyz^2 +\frac{x^4}{4}z \right) \bigg|_0^2 dy dz \\ \amp = \int_{-1}^1 \int_{-3}^3 \left( 8 + 6yz^2 +4z \right) dy dz = \int_{-1}^1 \left( 8y + 6\frac{y^3}{3}z^2 +4yz \right) \bigg|_{-3}^{3} dz \\ \amp = \int_{-1}^1 \left( 8(3-(-3)) + 2(27-(-3))z^2 +4(3-(-3))z \right) dz \\ \amp = \int_{-1}^1 \left( 48 + 60z^2 +24z \right) dz \\ \amp = 48z + 20z^3 +12z^2 \bigg|_{-1}^1 = 48(1-(-1)) + 20(1 - (-1)) + 12 (1-1) = 136 \end{align*}

Activity 6.5.9.

Integrate the function \(f(x,y) = \frac{1}{x-y}\) on the interval \([0,2] \times [1,3]\text{.}\)
Solution.
The function is undefined on the line \(y=x\text{.}\) That line passes through this interval, from \((1,1)\) to \((2,2)\text{.}\) Therefore, the integral of this function cannot be defined on this interval.

Activity 6.5.10.

Integrate the function \(f(x,y) = \sqrt{x^2 - y^2}\) on the interval \([0,1] \times [0,1]\text{.}\)
Solution.
The square root needs to be positive for this function to be defined. This happens when \(x^2 \geq y^2\text{,}\) which is two regions: \(x \geq y\) and \(-x \leq -y\) for positive \(x\) and \(y\text{.}\) On half of this region, \(x\) and \(y\) are both positive, but \(y > x\text{,}\) so the function is undefined and the integral is meaningless.

Activity 6.5.11.

Integrate the function \(f(x,y) = \frac{x^2y^2}{\sqrt{4 - x^2}}\) on the interval \([0,1] \times [-1,0]\text{.}\)
Solution.
This is an interated integral. I choose to integrate with \(x\) in the inside and \(y\) on the outside. Then the integral proceeds as two consecutive single-variable integrals. The antiderivative for the \(x\) integral is unweildy -- I asked a computer for the solution.
\begin{align*} \amp \int_0^1 \int_{-1}^0 \frac{x^2y^2}{\sqrt{4-x^2}} dx dy\\ \amp = \int_0^1 \frac{x^2}{\sqrt{4-x^2}} dx \int_{-1}^0 y^2 dy\\ \amp = \left[ 2 \arcsin \left( \frac{x}{2} \right) - \frac{1}{2} x \sqrt{4-x^2} \bigg|_0^1 \right] \left[ \frac{y^3}{3} \bigg|_{-1}^0 \right] \\ \amp = \left[ 2 \arcsin \frac{1}{2} - \frac{1}{2} \sqrt{3} - 2 \arcsin 0) + \frac{1}{2} (0) \sqrt{4} \right] \left[ 0 - \frac{-1}{3} \right] \\ \amp = \left[ 2 \frac{\pi}{6} \frac{1}{2} - \frac{\sqrt{3}}{2} - 0 + 0 \right] \left[ \frac{1}{3} \right] = \frac{\pi}{18} - \frac{\sqrt{3}}{6} \end{align*}

Activity 6.5.12.

Integrate the function \(f(x,y) = \frac{1}{(x + y)(x + 2y)(x + 3y)}\) on the interval \([-1,1] \times [2,4]\text{.}\)
Solution.
This is an interated integral. I choose to integrate with \(x\) in the inside and \(y\) on the outside. Then the integral proceeds as two consecutive single-variable integrals. The integral is well defined, since none of the line that solve either of the three factors in the denominator pass through the interval in question. In the first step for \(x\text{,}\) I need to use partial fractions to break up the integrand.
\begin{align*} \amp \int_2^4 \int_{-1}^1 \frac{1}{(x+y)(x+2y)(x+3y)} dx dy\\ \amp = \int_2^4 \int_{-1}^1 \frac{\frac{1}{2y^2}}{x+y} + \frac{\frac{-1}{y^2}}{x + 2y} + \frac{\frac{1}{2y^2}}{x + 3y} dx dy\\ \amp = \int_2^4 \frac{1}{2y^2} \ln |x+y| - \frac{-}{y^2} \ln |x+2y| + \frac{1}{2y^2} \ln |x+3y| \bigg|_{-1}{1} dy\\ \amp = \int_2^4 \frac{1}{2y^2} (\ln |1+y| - \ln |-1+y|) - \frac{1}{y^2} (\ln |1+2y| - \ln |-1+2y|) \\ \amp + \frac{1}{2y^2} (\ln |1+3y| - \ln |-1+3y|) dy \end{align*}
I can split this up into six different integrals, all of which are annoying to do. Integration by parts in each integral can remove the logarithmic part, leaving a rational function. Partial fractions can then deal with the rational functions. I skipped the details here, giving just the six integrals and their values calculated by a computer algebra system. Turns out it was a reasonable thing that I did so, since the numbers get quite miserable.
\begin{align*} \amp = \int_2^4 \frac{1}{2y^2} \ln |1+y| - \int_2^4 \frac{1}{2y^2} \ln |-1+y| - \int_2^4 \frac{1}{y^2} \ln |1+2y| \\ \amp + \int_2^4 \frac{1}{y^2} \ln |-1+2y| + \int_2^4 \frac{1}{2y^2} \ln |1+3y| - \int_2^4 \frac{1}{2y^2} \ln |-1+3y| \\ \amp = \frac{1}{8} \ln \frac{256}{3125} - \frac{1}{4} \ln \frac{4}{27} + \frac{1}{8} \ln \frac{27}{256} + \frac{\ln 2}{2} + \ln \frac{100}{81} \\ \amp + \frac{1}{2} \ln \frac{5}{3} + \frac{1}{2} \ln 3 + \ln \frac{49}{36} - \frac{1}{4} \ln 7 + \frac{1}{8} \ln \frac{1677216}{13^{13}} - \frac{1}{4} \ln \frac{64}{823543} \end{align*}

Subsection 6.5.3 Conceptual Review Questions

  • What is topology and why does it matter?
  • What are open and closed intervals in higher dimensions?
  • Why doesn’t the indefinite integral extend to higher dimensions?
  • What does it mean to measure the (hyper)volume under the graph of a scalar field?
  • What is an iterated integral?
  • Why can multivariable integral be split into single variable pieces?
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