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Worksheet Assigned Problems for Section 15.3

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.

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The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.

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Exercise Group.

In the following exercises, evaluate \(\ds \iiint_\c{B} f(x,y,z)\, dV\) for the specified function \(f\) and box \(\c{B}\text{.}\)

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15.3.3.

\(f(x,y,z) = xe^{y - 2z}\, ; \qquad 0 \leq x \leq 2\, , \quad 0 \leq y \leq 1\, , \quad 0 \leq z \leq 1\)

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Solution.

We are evaluating \(\ds \iiint_\c{B} xe^{y-2z} \, dV\) over the rectangular box \(\c{B} = [0,2] \times [0,1] \times [0,1]\text{.}\) Since all limits of integration are constants, we can set up the iterated integral in the order \(dz\, dy\, dx\text{:}\)

\begin{gather*} \int_0^2 \int_0^1 \int_0^1 xe^{y-2z} \, dz \, dy \, dx \end{gather*}

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Evaluate the inner integral with respect to \(z\text{:}\)

\begin{align*} \int_0^1 xe^{y-2z} \, dz \amp= \left[ -\frac{1}{2}xe^{y-2z} \right]_{z=0}^{z=1} \\ \amp= -\frac{1}{2}xe^{y-2} - \left(-\frac{1}{2}xe^{y-0}\right) \\ \amp= -\frac{1}{2}xe^{y-2} + \frac{1}{2}xe^y \\ \amp= \frac{1}{2}xe^y(1 - e^{-2}) \end{align*}

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Substitute this into the middle integral and evaluate with respect to \(y\text{:}\)

\begin{align*} \int_0^1 \frac{1}{2}xe^y(1 - e^{-2}) \, dy \amp= \frac{1}{2}x(1 - e^{-2}) \int_0^1 e^y \, dy \\ \amp= \frac{1}{2}x(1 - e^{-2}) \Big[ e^y \Big]_0^1 \\ \amp= \frac{1}{2}x(1 - e^{-2})(e - 1) \end{align*}

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Finally, substitute this into the outer integral and evaluate with respect to \(x\text{:}\)

\begin{align*} \int_0^2 \frac{1}{2}x(1 - e^{-2})(e - 1) \, dx \amp= \frac{1}{2}(1 - e^{-2})(e - 1) \int_0^2 x \, dx \\ \amp= \frac{1}{2}(1 - e^{-2})(e - 1) \left[ \frac{1}{2}x^2 \right]_0^2 \\ \amp= \frac{1}{2}(1 - e^{-2})(e - 1) (2 - 0) \\ \amp= (1 - e^{-2})(e - 1) \end{align*}

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15.3.7.

\(f(x,y,z) = (x + z)^3\, ; \qquad [0,a] \times [0,b] \times [0,c]\)

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Solution.

We are evaluating \(\ds \iiint_\c{B} (x+z)^3 \, dV\) over the rectangular box \(\c{B} = [0,a] \times [0,b] \times [0,c]\text{.}\) Let’s set up the iterated integral in the order \(dz\, dy\, dx\text{:}\)

\begin{gather*} \int_0^a \int_0^b \int_0^c (x+z)^3 \, dz \, dy \, dx \end{gather*}

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Evaluate the inner integral with respect to \(z\text{:}\)

\begin{align*} \int_0^c (x+z)^3 \, dz \amp= \left[ \frac{1}{4}(x+z)^4 \right]_{z=0}^{z=c} \\ \amp= \frac{1}{4}(x+c)^4 - \frac{1}{4}(x+0)^4 \\ \amp= \frac{1}{4}\big((x+c)^4 - x^4\big) \end{align*}

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Substitute this into the middle integral and evaluate with respect to \(y\text{:}\)

\begin{align*} \int_0^b \frac{1}{4}\big((x+c)^4 - x^4\big) \, dy \amp= \frac{1}{4}\big((x+c)^4 - x^4\big) \Big[ y \Big]_0^b \\ \amp= \frac{b}{4}\big((x+c)^4 - x^4\big) \end{align*}

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Substitute this into the outer integral and evaluate with respect to \(x\text{:}\)

\begin{align*} \int_0^a \frac{b}{4}\big((x+c)^4 - x^4\big) \, dx \amp= \frac{b}{4} \left[ \frac{1}{5}(x+c)^5 - \frac{1}{5}x^5 \right]_0^a \\ \amp= \frac{b}{20} \Big( \big((a+c)^5 - a^5\big) - \big((0+c)^5 - 0^5\big) \Big) \\ \amp= \frac{b}{20} \big( (a+c)^5 - a^5 - c^5 \big) \end{align*}

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Exercise Group.

In the following exercises, evaluate \(\ds \iiint_\c{W} f(x,y,z)\, dV\) for the function \(f\) and region \(\c{W}\) specified.

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15.3.9.

\(f(x,y,z) = x + y\, ; \qquad \c{W}: \, y \leq z \leq x\, , \quad 0 \leq y \leq x\, , \quad 0 \leq x \leq 1\)

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Solution.

The region \(\c{W}\) is already described perfectly for an iterated integral in the order \(dz\, dy\, dx\text{.}\)

\begin{align*} \iiint_\c{W} (x+y) \, dV \amp= \int_0^1 \int_0^x \int_y^x (x+y) \, dz \, dy \, dx \end{align*}

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Evaluate the inner integral with respect to \(z\text{:}\)

\begin{align*} \int_y^x (x+y) \, dz \amp= \Big[ (x+y)z \Big]_{z=y}^{z=x} \\ \amp= (x+y)(x) - (x+y)(y) \\ \amp= (x+y)(x-y) = x^2 - y^2 \end{align*}

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Substitute this into the middle integral and evaluate with respect to \(y\text{:}\)

\begin{align*} \int_0^x (x^2 - y^2) \, dy \amp= \left[ x^2y - \frac{1}{3}y^3 \right]_0^x \\ \amp= \left( x^3 - \frac{1}{3}x^3 \right) - (0) = \frac{2}{3}x^3 \end{align*}

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Substitute this into the outer integral and evaluate with respect to \(x\text{:}\)

\begin{align*} \int_0^1 \frac{2}{3}x^3 \, dx \amp= \left[ \frac{2}{12}x^4 \right]_0^1 = \left[ \frac{1}{6}x^4 \right]_0^1 \\ \amp= \frac{1}{6}(1) - 0 = \frac{1}{6} \end{align*}

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15.3.13.

\(f(x,y,z) = e^z\, ; \qquad \c{W}: \, x + y + z \leq 1\, , \quad x \geq 0\, , \quad y \geq 0\, , \quad z \geq 0\)

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Solution.

The region \(\c{W}\) is the tetrahedron in the first octant bounded by the plane \(x + y + z = 1\) and the three coordinate planes. We can set this up as a \(z\)-simple region. The top surface is the plane \(z = 1 - x - y\text{,}\) and the bottom surface is the \(xy\)-plane (\(z = 0\)).

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The projection of this solid onto the \(xy\)-plane is the triangle bounded by \(x = 0\text{,}\) \(y = 0\text{,}\) and the line \(x + y = 1\) (which comes from setting \(z = 0\)). For the outer limits, \(x\) goes from \(0\) to \(1\text{.}\) For a fixed \(x\text{,}\) \(y\) goes from \(0\) to \(1 - x\text{.}\)

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Setting up the iterated integral:

\begin{align*} \iiint_\c{W} e^z \, dV \amp= \int_0^1 \int_0^{1-x} \int_0^{1-x-y} e^z \, dz \, dy \, dx \end{align*}

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Evaluate the inner integral with respect to \(z\text{:}\)

\begin{align*} \int_0^{1-x-y} e^z \, dz \amp= \Big[ e^z \Big]_{z=0}^{z=1-x-y} \\ \amp= e^{1-x-y} - e^0 = e^{1-x-y} - 1 \end{align*}

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Evaluate the middle integral with respect to \(y\text{:}\)

\begin{align*} \int_0^{1-x} (e^{1-x-y} - 1) \, dy \amp= \Big[ -e^{1-x-y} - y \Big]_{y=0}^{y=1-x} \\ \amp= \Big( -e^{1-x-(1-x)} - (1-x) \Big) - \Big( -e^{1-x-0} - 0 \Big) \\ \amp= \Big( -e^0 - 1 + x \Big) - \Big( -e^{1-x} \Big) \\ \amp= -1 - 1 + x + e^{1-x} = x - 2 + e^{1-x} \end{align*}

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Finally, evaluate the outer integral with respect to \(x\text{:}\)

\begin{align*} \int_0^1 (x - 2 + e^{1-x}) \, dx \amp= \left[ \frac{1}{2}x^2 - 2x - e^{1-x} \right]_0^1 \\ \amp= \left( \frac{1}{2}(1)^2 - 2(1) - e^0 \right) - \left( 0 - 0 - e^1 \right) \\ \amp= \left( \frac{1}{2} - 2 - 1 \right) - (-e) \\ \amp= -\frac{5}{2} + e = e - \frac{5}{2} \end{align*}

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15.3.17.

Integrate \(f(x,y,z) = x\) over the region in the first octant bounded above by \(z = 8 - 2x^2 - y^2\) and below by \(z = y^2\text{.}\)

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Solution.

The region is bounded above by the paraboloid \(z = 8 - 2x^2 - y^2\) and below by the parabolic cylinder \(z = y^2\text{.}\) We are restricted to the first octant, which means \(x \geq 0, y \geq 0, z \geq 0\text{.}\)

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To find the domain \(\c{D}\) in the \(xy\)-plane, we find the curve where the top and bottom surfaces intersect:

\begin{align*} 8 - 2x^2 - y^2 \amp= y^2 \\ 8 \amp= 2x^2 + 2y^2 \\ x^2 + y^2 \amp= 4 \end{align*}

Since we are in the first octant, \(\c{D}\) is the quarter-circle of radius 2 in the first quadrant. As a \(y\)-simple region, \(x\) goes from \(0\) to \(2\text{,}\) and \(y\) goes from \(0\) to \(\sqrt{4-x^2}\text{.}\)

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Setting up the iterated integral:

\begin{align*} \iiint_\c{W} x \, dV \amp= \int_0^2 \int_0^{\sqrt{4-x^2}} \int_{y^2}^{8-2x^2-y^2} x \, dz \, dy \, dx \end{align*}

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Evaluate the inner integral with respect to \(z\text{:}\)

\begin{align*} \int_{y^2}^{8-2x^2-y^2} x \, dz \amp= \Big[ xz \Big]_{z=y^2}^{z=8-2x^2-y^2} \\ \amp= x(8 - 2x^2 - y^2) - xy^2 \\ \amp= x(8 - 2x^2 - 2y^2) = 8x - 2x^3 - 2xy^2 \end{align*}

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Substitute this into the middle integral and evaluate with respect to \(y\text{:}\)

\begin{align*} \int_0^{\sqrt{4-x^2}} (8x - 2x^3 - 2xy^2) \, dy \amp= \left[ (8x - 2x^3)y - \frac{2}{3}xy^3 \right]_0^{\sqrt{4-x^2}} \\ \amp= (8x - 2x^3)\sqrt{4-x^2} - \frac{2}{3}x(\sqrt{4-x^2})^3 - 0 \\ \amp= 2x(4 - x^2)\sqrt{4-x^2} - \frac{2}{3}x(4-x^2)^{3/2} \\ \amp= 2x(4 - x^2)^{3/2} - \frac{2}{3}x(4-x^2)^{3/2} \\ \amp= \frac{4}{3}x(4 - x^2)^{3/2} \end{align*}

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Substitute this into the outer integral and evaluate with respect to \(x\) using a \(u\)-substitution. Let \(u = 4 - x^2\text{,}\) then \(du = -2x \, dx\text{.}\) When \(x=0, u=4\) and when \(x=2, u=0\text{.}\)

\begin{align*} \int_0^2 \frac{4}{3}x(4 - x^2)^{3/2} \, dx \amp= \int_4^0 \frac{4}{3} \left(-\frac{1}{2}\right) u^{3/2} \, du \\ \amp= -\frac{2}{3} \int_4^0 u^{3/2} \, du \\ \amp= \frac{2}{3} \int_0^4 u^{3/2} \, du \\ \amp= \frac{2}{3} \left[ \frac{2}{5}u^{5/2} \right]_0^4 \\ \amp= \frac{4}{15} \left( 4^{5/2} - 0 \right) \\ \amp= \frac{4}{15} (32) = \frac{128}{15} \end{align*}

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15.3.21.

Find the volume of the solid in the first octant bounded between the planes \(x + y + z = 1\) and \(x + y + 2z = 1\text{.}\)

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Solution.

We want to find the volume, which means our objective function is simply \(f(x,y,z) = 1\text{.}\) The region is bounded between the two planes \(x + y + z = 1\) and \(x + y + 2z = 1\) in the first octant (\(x \geq 0, y \geq 0, z \geq 0\)).

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Solving both planes for \(z\text{:}\)

Plane 1: \(z = 1 - x - y\)

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Plane 2: \(z = \frac{1}{2}(1 - x - y)\)

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Since \(x\) and \(y\) are positive in the first octant, \((1 - x - y)\) is larger than \(\frac{1}{2}(1 - x - y)\text{.}\) Thus, Plane 1 is the "top" surface and Plane 2 is the "bottom" surface.

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These two planes intersect exactly when \(z = 0\text{.}\) If we set \(z = 0\text{,}\) both equations yield the line \(x + y = 1\text{.}\) This line, along with the axes, forms the triangular domain \(\c{D}\) in the \(xy\)-plane. As a vertically simple region, \(x\) goes from \(0\) to \(1\text{,}\) and \(y\) goes from \(0\) to \(1 - x\text{.}\)

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Setting up the integral for volume:

\begin{align*} V \amp= \int_0^1 \int_0^{1-x} \int_{\frac{1}{2}(1-x-y)}^{1-x-y} 1 \, dz \, dy \, dx \end{align*}

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Evaluate the inner integral:

\begin{align*} \int_{\frac{1}{2}(1-x-y)}^{1-x-y} 1 \, dz \amp= \Big[ z \Big]_{\frac{1}{2}(1-x-y)}^{1-x-y} \\ \amp= (1 - x - y) - \frac{1}{2}(1 - x - y) = \frac{1}{2}(1 - x - y) \end{align*}

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Evaluate the middle integral:

\begin{align*} \int_0^{1-x} \frac{1}{2}(1 - x - y) \, dy \amp= \frac{1}{2} \left[ y - xy - \frac{1}{2}y^2 \right]_0^{1-x} \\ \amp= \frac{1}{2} \left( (1-x) - x(1-x) - \frac{1}{2}(1-x)^2 \right) - 0 \\ \amp= \frac{1}{2} \left( (1-x) - x + x^2 - \frac{1}{2}(1 - 2x + x^2) \right) \\ \amp= \frac{1}{2} \left( 1 - 2x + x^2 - \frac{1}{2} + x - \frac{1}{2}x^2 \right) \\ \amp= \frac{1}{2} \left( \frac{1}{2}x^2 - x + \frac{1}{2} \right) = \frac{1}{4}(x^2 - 2x + 1) = \frac{1}{4}(x - 1)^2 \end{align*}

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Evaluate the outer integral:

\begin{align*} \int_0^1 \frac{1}{4}(x - 1)^2 \, dx \amp= \left[ \frac{1}{12}(x - 1)^3 \right]_0^1 \\ \amp= \frac{1}{12}(0) - \frac{1}{12}(-1)^3 = \frac{1}{12} \end{align*}

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15.3.25.

Describe the domain of integration of the following integral

\begin{equation*} \int_{-2}^2 \int_{-\sqrt{4 - z^2}}^{\sqrt{4 - z^2}} \int_1^{\sqrt{5 - x^2 - z^2}} f(x,y,z)\, dy \, dx \, dz \end{equation*}

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Solution.

Let’s look at the given limits of integration carefully:

\begin{align*} -2 \amp\leq z \leq 2 \\ -\sqrt{4-z^2} \amp\leq x \leq \sqrt{4-z^2} \\ 1 \amp\leq y \leq \sqrt{5 - x^2 - z^2} \end{align*}

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The inner limits tell us that \(y\) goes from the plane \(y = 1\) to the surface \(y = \sqrt{5 - x^2 - z^2}\text{.}\) If we square the top surface equation, we get \(y^2 = 5 - x^2 - z^2\text{,}\) or \(x^2 + y^2 + z^2 = 5\text{.}\) This is a sphere of radius \(\sqrt{5}\) centered at the origin. Since \(y\) is given by the positive square root, it is the "front" hemisphere. So, the region is bounded between the flat plane \(y = 1\) and the spherical cap \(x^2 + y^2 + z^2 = 5\text{.}\)

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The middle limits describe how \(x\) behaves: it goes from the left half of a circle \(x = -\sqrt{4-z^2}\) to the right half \(x = \sqrt{4-z^2}\text{.}\) Squaring this gives \(x^2 = 4 - z^2\) or \(x^2 + z^2 = 4\text{.}\) This tells us the shadow (projection) of the solid onto the \(xz\)-plane is exactly a disk of radius 2 centered at the origin.

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The outer limits confirm this, as \(z\) goes from \(-2\) to \(2\text{,}\) perfectly sweeping out that disk of radius 2.

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Conclusion: The domain of integration is the solid chunk of the sphere \(x^2 + y^2 + z^2 \leq 5\) that is sliced off by the plane \(y = 1\) and lies in the direction of the positive \(y\)-axis (the "cap" of the sphere). Notice that if you plug \(y=1\) into the sphere equation, you get \(x^2 + 1 + z^2 = 5 \implies x^2 + z^2 = 4\text{,}\) which perfectly matches the projected disk on the \(xz\)-plane!

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15.3.31.

Let \(\ds \c{W} = \left\{(x,y,z) \mid \sqrt{x^2 + y^2} \leq z \leq 1 \right\}\) (see the figure below). Express \(\ds \iiint_\c{W} f(x,y,z)\, dV\) as an iterated integral in the order \(dz\, dy\, dx\) (for an arbitrary function \(f\)).

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Solution.

The region \(\c{W}\) is bounded below by the cone \(z = \sqrt{x^2 + y^2}\) and above by the flat plane \(z = 1\text{.}\) This describes a solid cone that opens upwards, with its tip at the origin and a flat circular top at height 1.

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To express the integral in the order \(dz\, dy\, dx\text{,}\) we first need the bounds for \(z\text{.}\) From the problem description, \(z\) goes from the cone to the plane:

\begin{gather*} \sqrt{x^2 + y^2} \leq z \leq 1 \end{gather*}

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Next, we find the projection \(\c{D}\) of this solid onto the \(xy\)-plane to get the bounds for \(y\) and \(x\text{.}\) The widest part of this solid is at the top where \(z = 1\text{.}\) Substituting \(z = 1\) into the cone equation gives:

\begin{gather*} 1 = \sqrt{x^2 + y^2} \implies x^2 + y^2 = 1 \end{gather*}

So the projection \(\c{D}\) is the unit disk \(x^2 + y^2 \leq 1\text{.}\)

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To set up the \(dy\, dx\) outer integrals for a unit disk, \(x\) goes from \(-1\) to \(1\text{.}\) For a fixed \(x\text{,}\) \(y\) goes from the bottom semicircle to the top semicircle.

\begin{align*} -1 \amp\leq x \leq 1 \\ -\sqrt{1-x^2} \amp\leq y \leq \sqrt{1-x^2} \end{align*}

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Putting it all together, the iterated integral is:

\begin{gather*} \iiint_\c{W} f(x,y,z) \, dV = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{\sqrt{x^2+y^2}}^1 f(x,y,z) \, dz \, dy \, dx \end{gather*}

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15.3.37.

Draw the region \(\c{W}\) bounded by the surfaces given by \(z = y^2\text{,}\) \(y = z^2\text{,}\) and the planes given by \(x = 0\text{,}\) \(x + y + z = 4\text{.}\) Then set up (but do not compute) a single triple integral that yields the volume of \(\c{W}\text{.}\)

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Solution.

Let’s analyze the bounding surfaces to understand the region \(\c{W}\text{:}\)

\(z = y^2\text{:}\) A parabolic cylinder opening upward along the \(z\)-axis.

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\(y = z^2\text{:}\) A parabolic cylinder opening forward along the \(y\)-axis.

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\(x = 0\text{:}\) The \(yz\)-plane (a back wall).

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\(x + y + z = 4 \implies x = 4 - y - z\text{:}\) A slanted plane acting as a front wall.

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If we look closely at the first two surfaces, \(z = y^2\) and \(y = z^2\text{,}\) they intersect to form a tube-like region that extends infinitely along the \(x\)-axis. In the \(yz\)-plane, these two parabolas intersect at \((0,0)\) and \((1,1)\text{,}\) creating a small leaf-shaped region.

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Because the tube runs along the \(x\)-axis, it makes sense to treat this as an \(x\)-simple region (integrating with respect to \(x\) first). The back wall is \(x = 0\) and the front wall is the slanted plane \(x = 4 - y - z\text{.}\)

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The projection \(\c{D}\) onto the \(yz\)-plane is exactly that leaf-shaped region bounded by the two parabolas. Setting them up in a \(dz\, dy\) order: \(y\) ranges from 0 to 1. For a fixed \(y\text{,}\) \(z\) goes from the lower parabola \(z = y^2\) to the upper parabola (which we must solve for \(z\text{:}\) \(y = z^2 \implies z = \sqrt{y}\)).

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Since we want the volume, the objective function is \(f(x,y,z) = 1\text{.}\) The triple integral setup is:

\begin{gather*} V = \int_0^1 \int_{y^2}^{\sqrt{y}} \int_0^{4-y-z} 1 \, dx \, dz \, dy \end{gather*}

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