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Section 1.4 Week 1 Activity

Subsection 1.4.1 Iterated Integrals

Activity 1.4.1.

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 1.4.2.

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 1.4.3.

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 1.4.4.

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 1.4.5.

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 1.4.6.

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 1.4.7.

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 1.4.8.

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 1.4.9.

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 1.4.2 Conceptual Review Questions

  • 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?