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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 7.4 Week 7 Activity

Subsection 7.4.1 Activity

Activity 7.4.1.

Integrate the function \(f(x,y) = \frac{xy}{5} + \frac{y^2}{6} \) on the triangle with vertices \((0,0)\text{,}\) \((4,0)\) and \((4,3)\text{.}\)
Solution.
The line forming the longest side of the triangle is the line \(y = \frac{3}{4}x\text{.}\) If I take \(x \in [0,4]\) then \(y\) goes from the \(x\) axis (\(y=0\)) to this line. I set up the integral with constant bounds in \(x\) and variable bounds in \(y\text{.}\)
\begin{align*} \amp \int_0^4 \int_0^{\frac{3x}{4}} \frac{xy}{5} + \frac{y^2}{6} dy dx\\ \amp = \int_0^4 \frac{xy^2}{10} + \frac{y^3}{18} \bigg|_0^{\frac{3x}{4}} dx \\ \amp = \int_0^4 \frac{x}{10} \left( \frac{9x^2}{16} - 0 \right) + \frac{1}{18} \left( \frac{27x^3}{64} - 0 \right) dx \\ \amp = \int_0^4 \frac{9x^3}{160} + \frac{27x^3}{1152} dx\\ \amp = \frac{51}{640} \int_0^4 x^3 dx = \frac{51}{640} \frac{x^4}{4} \bigg|_0^4 = \frac{51}{640} \frac{256}{4} = \frac{51}{10} \end{align*}

Activity 7.4.2.

Integrate the function \(f(x,y) = x^2y^2 - x + y\) on the parallelogram with vertices \((0,0)\text{,}\) \((0,2)\text{,}\) \((2,2)\) and \((2,4)\text{.}\)
Solution.
The lower line of the parallelogram is \(y=x\text{.}\) The higher line is \(y = x + 2\text{.}\) If I let \(x \in [0,2]\text{,}\) then I can let \(y\) go between these two lines. I set up an integral with constant bounds in \(x\) and variable bound sin \(y\text{.}\)
\begin{align*} \amp \int_0^2 \int_{x}^{x+2} x^2y^2 - x + y dy dx = \int_0^2 \frac{x^2y^3}{3} - xy + \frac{y^2}{2} \bigg|_{x}^{x+2} dx \\ \amp = \int_0^2 \frac{x^2}{3} \left( (x+2)^3 - (x)^3 \right) - x((x+2) - x) + \frac{1}{2} \left( (x+2)^2 - (x)^2 \right) dx \\ \amp = \int_0^2 \frac{x^2}{3} (x^3 + 6x^2 + 12x + 8 - x^3) - 2x + \frac{1}{2} (x^2 + 4x + 4 - x^2) dx\\ \amp = \int_0^2 2x^4 + 4x^3 + \frac{8x^2}{3} - 2x + 2x + 2 dx = \int_0^2 2x^4 + 4x^3 + \frac{8x^2}{3} + 2 dx \\ \amp = \frac{2x^5}{5} + x^4 + \frac{8x^3}{9} + 2x \bigg|_0^2\\ \amp = \frac{64}{5} + 16 + \frac{64}{9} + 4 = \frac{1796}{45} \end{align*}

Activity 7.4.3.

Integrate the function \(f(x,y) = xy\) on the halfcircle with radius \(3\) above the \(x\) axis.
Solution.
If I let \(x \in [-3,3]\text{,}\) then the halfcircle is bounded between the \(x\) axis (\(y=0\)) and \(y = \sqrt{9-x^2}\text{.}\) I have the positive square root, since this is the half-circle above the \(x\) axis. I setup the integral with constant bounds in \(x\) and variable bounds in \(y\text{.}\)
\begin{align*} \amp \int_{-3}^3 \int_0^{\sqrt{9-x^2}} xy dx \\ \amp = \int_{-3}^3 \frac{xy^2}{2} \bigg|_{\sqrt{9-x^2}} dx = \int_{-3}^3 \frac{x}{2} (\sqrt{9-x^2})^2 - 0 dx = \int_{-3}^3 \frac{x}{2} (9-x^2) dx \\ \amp = \frac{1}{2} \int_{-3}^3 9x - x^3 dx = \frac{1}{2} \left( \frac{9x^2}{2} - \frac{x^4}{4} \bigg|_{-3}^3 \right) \\ \amp = \frac{1}{2} \left( \frac{9}{2} (3^x - (-3)^2) - \frac{1}{4} (3^4 - (-3)^4) \right) = 0 \end{align*}

Activity 7.4.4.

Integrate the function \(f(x,y) = e^{x+y}\) on the region between \(g(x) = \frac{x}{4}\) and \(h(x) = \frac{x}{3}\) for \(x \in [0,4]\text{.}\)
Solution.
I am given a constnat range for \(x\) and curves for \(y\) in terms of \(x\text{,}\) so I can use those to set up the integral. Note that I can use the laws of exponents to write this integrand in an easier form.
\begin{align*} \amp \int_0^4 \int_{\frac{x}{4}}^{\frac{x}{3}} e^{x+y} dy dx\\ \amp = \int_0^4 \int_{\frac{x}{4}}^{\frac{x}{3}} e^x e^y dy dx = \int_0^4 e^x dx \int_{\frac{x}{4}}^{\frac{x}{3}} e^y dy = \int_0^4 e^x e^y \bigg|_{\frac{x}{4}}^{\frac{x}{3}} dx\\ \amp = \int_0^4 e^x (e^\frac{x}{3} - e^{\frac{x}{4}} ) dx = \int_0^4 e^\frac{5x}{3} - e^{\frac{5x}{4}} ) dx = \frac{3}{5} e^\frac{5x}{3} - \frac{4}{5}e^{5x}{4} \bigg|_0^4\\ \amp = \frac{3}{5} (e^{\frac{20}{3}} - e^0) - \frac{4}{5} (e^5 - e^0 ) = \frac{3}{5} (e^{\frac{20}{3}} - 1) - \frac{4}{5} (e^5 - 1) \end{align*}

Activity 7.4.5.

Integrate the function \(f(x,y) = x+y\) on the crescent formed between \(y^2 = 4-x\) and \(y^2 = 8-4x\text{.}\)
Solution.
To find the region of integration, I need to find the intersections of the two parabolas. If I substitute for \(y^2\text{,}\) I get \(4-x = 8-4x\text{,}\) which is solved by \(x = \frac{4}{3}\text{.}\) The \(y\) coordinate, from either equation, is \(\pm \sqrt{\frac{8}{3}}\text{.}\) This gives me constant bounds for the \(y\) coordinate: \(\left[ -\sqrt{\frac{8}{3}}, \sqrt{\frac{8}{3}} \right]\text{.}\) If I use constant bounds for \(y\) in the integral, I can use \(x = 4-y^2\) and \(x = 2 - \frac{y^2}{4}\) as the bounds on \(x\text{.}\) The second curve has the smaller \(x\) valuse in this range, so it is the lower bound.
\begin{align*} \amp \int_{-\sqrt{\frac{8}{3}}}^{\sqrt{\frac{8}{3}}} \int_{2-\frac{y^2}{4}}^{4-y^2} x+y dx dy = \int_{-\sqrt{\frac{8}{3}}}^{\sqrt{\frac{8}{3}}} \frac{x^2}{2} + xy \bigg|_{2-\frac{y^2}{4}}^{4-y^2} dy\\ \amp = \int_{-\sqrt{\frac{8}{3}}}^{\sqrt{\frac{8}{3}}} \frac{1}{2} \left( (4-y^2)^2 - \left(2 - \frac{y^2}{4} \right) \right) + y \left( 4 - y^2 - \left( 2- \frac{y^2}{4} \right) \right) dy\\ \amp = \int_{-\sqrt{\frac{8}{3}}}^{\sqrt{\frac{8}{3}}} \frac{1}{2} \left( 16 - 8y^2 + y^4 - 4 + \frac{y^2}{4} - \frac{y^4}{16} \right) + 4y - y^3 - 2y + \frac{y^3}{4} dy\\ \amp = \int_{-\sqrt{\frac{8}{3}}}^{\sqrt{\frac{8}{3}}} 8 - 4y^2 + \frac{y^4}{2} - 2 + \frac{y^2}{8} - \frac{y^4}{32} + 2y - \frac{3y^3}{4} dy\\ \amp = \int_{-\sqrt{\frac{8}{3}}}^{\sqrt{\frac{8}{3}}} 6 + 2y - \frac{31y^2}{8} - \frac{3y^3}{4} + \frac{15y^4}{32} dy = 6y + y^2 - \frac{31y^3}{24} - 3y^4 + \frac{15y^5}{160} \bigg|_{-\sqrt{\frac{8}{3}}}^{\sqrt{\frac{8}{3}}}\\ \amp = 12 \sqrt{\frac{8}{3}} + 0 - \frac{31}{24} 2 \frac{8}{3} \sqrt{\frac{8}{3}} - 0 + \frac{15}{160} 2 \frac{64}{9} \sqrt{\frac{8}{3}} = \frac{182}{9} \sqrt{\frac{8}{3}}. \end{align*}

Activity 7.4.6.

Integrate the function \(f(x,y) = \sin (xy)\) on the annulus with inner radius \(2\) and outer radius \(4\text{.}\)
Solution.
I cannot approach the whole annulus as one regions with fixed bounds in \(x\) or \(y\) and variable bounds in the other. However, I can make us of some symmetry. Sine is an odd function, so the value on the second quadrant should be the negative of the value on the first quadrant. The same is true on the third and fourth quadrants. All in all, the various pieces of the integral will cancel out and the integral will evaluate to zero. I admit, this is a bit of a tricky answer, since I’m getting out ot actually describing the annulus with bounds. But it is worth remembering that these integrals are (hyper)volumes and sometimes there are good geometric arguments to be made that help to avoid calculations.

Activity 7.4.7.

Integrate the function \(f(x,y,z) = xy + z\) on the triangular pyramid with vertices \((0,0,0)\text{,}\) \((3,0,0)\text{,}\) \((0,2,0)\) and \((0,0,4)\text{.}\)
Solution.
First consider the face of the pyramid in the \(xy\) plane, which is a triangle. The longest edge of this face is part of the line \(y = 2-\frac{2}{3} x\text{.}\) I can start with \(x \in [0,3]\) and then set \(y\) between \(x=0\) and \(2 - \frac{2}{3}\) to describe bounds for this triangle. Then consider the tilted face of the pyramid -- I need the equation for this plane, since \(z\) will start on at \(z=0\) and go to this equation of this plane.
The vectors \((3,0,-4)\) and \((0,2,-4)\) are local directions on the plane. Their cross product is \((-8, -12, -6)\text{,}\) which I can scale to \((4,6,3)\text{.}\) Using any point, I find the equation of the plane is \(4x + 6y + 3z = 12\text{.}\) If I solve for \(z\text{,}\) I get \(z = 4 - \frac{4x}{3} - 2y\text{.}\) Therefore, I can take the bounds for \(z\) to be from \(z=0\) to \(z = 4 - \frac{4x}{3} - 2y\text{.}\)
\begin{align*} \amp \int_0^3 \int_0^{2 - \frac{2x}{3}} \int_0^{4 - \frac{4x}{3} - 2y} xy + z dz dy dx = \int_0^3 \int_0^{2 - \frac{2x}{3}} xyz + \frac{z^2}{2} \bigg|_0^{4 - \frac{4x}{3} -2y} dy dx\\ \amp = \int_0^3 \int_0^{2 - \frac{2x}{3}} xy \left( 4 - \frac{4x}{3} - 2y \right) + \frac{1}{2} \left( 4 - \frac{4x}{3} - 2y \right)^2 dy dx\\ \amp = \int_0^3 \int_0^{2 - \frac{2x}{3}} 4xy - \frac{4x^2y}{3} - 2xy^2 + 8 - \frac{16x}{3} - 8y + \frac{8yx}{3} + \frac{8x^2}{3} + 2y^2 dy dx\\ \amp = \int_0^3 \int_0^{2 - \frac{2x}{3}} 8 - \frac{16x}{3} - 8y + \frac{20xy}{3} + \frac{8x^2}{3} + 2y^2 - \frac{4x^2y}{3} -2xy^2 dy dx\\ \amp = \int_0^3 8y - \frac{16xy}{3} - 4y^2 + \frac{10xy^2}{3} + \frac{8x^2y}{3} + \frac{2y^3}{3} - \frac{2x^2y^2}{3} - \frac{2xy^3}{3} \bigg|_0^{2-\frac{2x}{3}} dx\\ \amp = \int_0^3 8 \left( 2 - \frac{2x}{3} \right) - \frac{16x}{3} \left( 2 - \frac{2x}{3} \right) - 4 \left( 2 - \frac{2x}{3} \right)^2 + \frac{10x}{3} \left( 2 - \frac{2x}{3} \right)^2 \\ \amp + \frac{8x^2}{3} \left( 2 - \frac{2x}{3} \right)^2 + \frac{2}{3} \left( 2 - \frac{2x}{3} \right)^3 - \frac{2x^2}{3} \left( 2 - \frac{2x}{3} \right)^2 - \frac{2x}{3} \left( 2 - \frac{2x}{3} \right)^3 dx\\ \amp = \int_0^3 8 \left( 2 - \frac{2x}{3} \right) - \frac{16x}{3} \left( 2 - \frac{2x}{3} \right) - 4 \left( 4 - \frac{8x}{3} + \frac{4x^2}{9} \right)\\ \amp + \frac{10x}{3} \left( 4 - \frac{8x}{3} + \frac{4x^2}{9} \right) + \frac{8x^2}{3} \left( 4 - \frac{8x}{3} + \frac{4x^2}{9} \right) + \frac{2}{3} \left( 8 - 12x + 12x^2 - \frac{8x^3}{27} \right) \\ \amp - \frac{2x^2}{3} \left( 4 - \frac{8x}{3} + \frac{4x^2}{9} \right) - \frac{2x}{3} \left( 8 - 12x + 12x^2 - \frac{8x^3}{27} \right) dx\\ \amp = \int_0^3 16 - \frac{16x}{3} - \frac{32}{3} + \frac{32x}{9} - 16 + \frac{32x}{3} - \frac{16x^2}{9} + \frac{40x}{3} - \frac{80x^2}{9} \\ \amp + \frac{40x^3}{27} + \frac{32x^2}{3} - \frac{64x^3}{9} + \frac{32x^4}{27} + \frac{16}{3} - 6x + 6x^2 \\ \amp - \frac{16x^3}{81} \frac{-8x^2}{3} + \frac{16x^3}{9} - \frac{8x^4}{27} \frac{-16x}{3} + 8x^2 - 8x^3 + \frac{16x^4}{81} dx \\ \amp = \int_0^3 \frac{-16}{3} - \frac{194x}{9} - \frac{16x^2}{9} - \frac{976x^3}{81} + \frac{88x^4}{81} \\ \amp = \frac{-16x}{3} - \frac{97x^2}{9} - \frac{16x^3}{27} - \frac{244x^4}{81} + \frac{88x^5}{405} \bigg|_0^3 \\ \amp = -16 + 97 - 16 - 244 + \frac{264}{5} = \frac{-631}{5} \end{align*}

Activity 7.4.8.

Integrate the function \(f(x,y,z) = x^2 + y^2 + z^2\) on the cone above the \(xy\) plane with radius \(3\) and height \(5\text{.}\)
Solution.
I need a description of the cone to use for the domain of the integral. In the \(x-y\) plane, the cone is a circle, so I can take \(x \in [-3,3]\) and \(y\) between the curves \(-\sqrt{9-x^2}\) and \(\sqrt{9-x^2}\text{.}\) For \(z\text{,}\) I need to start at \(z\) and stop at the cone, so I need the cone as the graph of a function of \(x\) and \(y\text{.}\) The function \(g(x,y) = 5 - \frac{5}{3}\sqrt{x^2+y^2}\) describes the cone.
\begin{align*} \amp \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_0^{5 - \frac{5}{3} \sqrt{x^2 +y^2}} xy(z-5) dz dy dx \\ \amp = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{xy(z-5)^2}{2} \bigg|_0^{5 - \frac{5}{3} \sqrt{x^2 +y^2}} xy(z-5) dy dx\\ \amp = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{xy}{2} \left( \left(\frac{-5}{3} \sqrt{x^2 +y^2} \right)^2 - (-5)2 \right) dy dx\\ \amp = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{xy}{2} \left( \frac{25(x^2+y^2)}{9} - 25 \right) dy dx \\ \amp = \frac{25}{18} \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \left(x^3y + xy^3 - xy \right) dy dx \end{align*}
I can split this into three integral by linearity and deal with each integral in turn.
\begin{align*} \amp \frac{25}{18} \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x^3y dy dx = \frac{25}{18} \int_{-3}^3 \frac{x^3y^2}{2} \bigg|_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} dx + \frac{25}{18} = 0 \\ \amp \frac{25}{18} \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} xy^3 dy dx = \int_{-3}^3 \frac{xy^4}{4} \bigg|_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} dx - \frac{25}{2} = 0 \\ \amp \frac{-25}{2} \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} xy dy dx = \int_{-3}^3 \frac{xy^2}{2} \bigg|_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} dx = 0 \end{align*}
All of these integrals evaluate to zero. In the evaluate of the bounds of the \(y\) integral, the \(y\) term is either \(y^2\) or \(y^4\) and the bounds are symmetric, positive and negative. When I put this into \(y^2\) or \(y^4\text{,}\) the negatives become positive, and the term becomes \((9-x^2) - (9-x^2)\) or \((9-x^2)^2 - (9-x^2)^2\text{.}\) In either case, this evaluates to zero, making the remaining integral in \(x\) just an integral of zero.

Activity 7.4.9.

The function \(f(x,y,z) = xyz\) on the region between \(z=x^2+y^2\) and \(z = 2(x^2+y^2)-3\text{.}\)
Solution.
I need to understand the intersection of these two surfaces. I can substitute for \(z\text{,}\) since it is already isolated, which lets me solve to find \(x^2 + y^2 = 3\text{.}\) In the first equation, that also gives \(z=3\text{.}\) So the intersection of the two surfaces is the circle of radius \(3\) in the \(z=3\) plane. I want the region between the two surfaces, which must be the region between the surfaces inside the radius of this circle. Inside the circle, \(x^2 + y^2 > 2(x^2+y^2) - 3\text{,}\) so \(z\) ranges from \(2(x^2+y^2)-3\) to \(x^2+y^2\) I can describe the region as \(x \in [-3,3]\text{,}\) \(y\) between \(-\sqrt{9-x^2}\) and \(\sqrt{9-x^2}\) and then \(z\) varies betwen the two surfaces. I’ve used a computer algebra system for expansions of some of the polynomial terms.
\begin{align*} \amp \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{2(x^2+y^2)-3}^{x^2+y^2} xyz dz dy dx\\ \amp = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{xyz^2}{2} \bigg|_{2(x^2+y^2)-3}^{x^2+y^2} dy dx \\ \amp = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{xy}{2} \left( (x^2+y^2)^2 - (2(x^2+y^2)-3)^2 \right) dy dz\\ \amp = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{-3x^5y}{2} - 3x^3y^3 + 6x^3y - \frac{3xy^5}{2} + 6 xy^3 - \frac{9xy}{2} dy dz \\ \amp = \int_{-3}^3 \frac{-3x^5y^2}{4} - \frac{3x^3y^4}{4} + 3x^3y^2 - \frac{xy^6}{4} + \frac{3xy^4}{2} - \frac{9xy^2}{4} \bigg|_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} dz = 0 \end{align*}
As with the previous question, I get all even powers of \(y\) to evaluate on these bounds. Since the bound differ by only a sign, the subtraction will cancel out and give zero.

Subsection 7.4.2 Conceptual Review Questions

  • How to variable bound describe different regions of integration?
  • Why does a bound involving a variable need to be inside the integral involving that variable?
  • How can we change order of integration with variable bounds? Why might it be useful to do so?
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