We need to determine the region \(R\) over which we will integrate. To do so, we need to determine where the planes intersect. They have common \(z\)-values when \(3x+y-4=8-3x-2y\text{.}\) Applying a little algebra, we have: \begin{align*} 3x+y-4 \amp = 8-3x-2y\\ 6x+3y \amp =12\\ 2x+y \amp =4 \end{align*}

The planes intersect along the line \(2x+y=4\text{.}\) Therefore the region \(R\) is bounded by \(x=0\text{,}\) \(y=0\text{,}\) and \(y=4-2x\text{;}\) we can convert these bounds to integration bounds of \(0\leq x\leq 2\text{,}\) \(0\leq y\leq 4-2x\text{.}\) Thus \begin{align*} V \amp = \iint_R \big(8-3x-2y-(3x+y-4)\big)\ dA\\ \amp = \int_0^2\int_0^{4-2x} \big(12-6x-3y\big)\ dy\ dx\\ \amp = 16\text{u} ^3. \end{align*}

The volume between the surfaces is \(16\) cubic units.

Starting with the order of integration \(dz\ dy\ dx\text{,}\) we need to first find bounds on \(z\text{.}\) The region \(D\) is bounded below by the plane \(z=0\) (because we are restricted to the first octant) and above by \(z=2-y/3-2x/3\text{;}\) \(0\leq z\leq 2-y/3-2x/3\text{.}\)

To find the bounds on \(y\) and \(x\text{,}\) we “collapse” the region onto the \(x\)-\(y\) plane, giving the triangle shown in Figure 13.6.12(b). (We know the equation of the line \(y=6-2x\) in two ways. First, by setting \(z=0\text{,}\) we have \(0 = 2-y/3-2x/3 \Rightarrow y=6-2x\text{.}\) Secondly, we know this is going to be a straight line between the points \((3,0)\) and \((0,6)\) in the \(x\)-\(y\) plane.)

We define that region \(R\text{,}\) in the integration order of \(dy\ dx\text{,}\) with bounds \(0\leq y\leq 6-2x\) and \(0\leq x\leq 3\text{.}\) Thus the volume \(V\) of the region \(D\) is: \begin{align*} V \amp = \iiint_D \ dV\\ \amp = \int_0^3\int_0^{6-2x}\int_0^{2-\frac 13y-\frac 23x}\ dz\ dy\ dz\\ \amp = \int_0^3\int_0^{6-2x}\left(\int_0^{2-\frac 13y-\frac 23x}\ dz\right)\ dy\ dz\\ \amp =\int_0^3\int_0^{6-2x}z\Big|_0^{2-\frac 13y-\frac 23x}\ dy\ dz\\ \amp = \int_0^3\int_0^{6-2x}\left(2-\frac 13y-\frac 23x\right)\ dy\ dz.\\ \end{align*} From this step on, we are evaluating a double integral as done many times before. We skip these steps and give the final volume, \begin{align*} \amp = 6\text{u} ^3. \end{align*}

The order \(dz\ dx\ dy\text{:}\)

Now consider the volume using the order of integration \(dz\ dx\ dy\text{.}\) The bounds on \(z\) are the same as before, \(0\leq z\leq 2-y/3-2x/3\text{.}\) Collapsing the space region on the \(x\)-\(y\) plane as shown in Figure 13.6.12(b), we now describe this triangle with the order of integration \(dx\ dy\text{.}\) This gives bounds \(0\leq x\leq 3-y/2\) and \(0\leq y\leq 6\text{.}\) Thus the volume is given by the triple integral \begin{equation*} V = \int_0^6\int_0^{3-\frac12y}\int_0^{2-\frac13y-\frac23x}\ dz\ dx\ dy. \end{equation*}

The order \(dx\ dy\ dz\text{:}\)

Following our “surface to surface\(\ldots\)” strategy, we need to determine the \(x\)-surfaces that bound our space region. To do so, approach the region “from behind,” in the direction of increasing \(x\text{.}\) The first surface we hit as we enter the region is the \(y\)-\(z\) plane, defined by \(x=0\text{.}\) We come out of the region at the plane \(z=2-y/3-2x/3\text{;}\) solving for \(x\text{,}\) we have \(x= 3-y/2-3z/2\text{.}\) Thus the bounds on \(x\) are: \(0\leq x\leq 3-y/2-3z/2\text{.}\)

Now collapse the space region onto the \(y\)-\(z\) plane, as shown in Figure 13.6.15(a). (Again, we find the equation of the line \(z=2-y/3\) by setting \(x=0\) in the equation \(x=3-y/2-3z/2\text{.}\)) We need to find bounds on this region with the order \(dy\ dz\text{.}\) The curves that bound \(y\) are \(y=0\) and \(y=6-3z\text{;}\) the points that bound \(z\) are 0 and 2. Thus the triple integral giving volume is: \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq x\leq 3-y/2-3z/2\\ 0\leq y\leq 6-3z\\ 0\leq z\leq 2 \end{array} \amp \ds\Rightarrow \int_0^2\int_0^{6-3z}\int_0^{3-y/2-3z/2}\ dx\ dy\ dz. \end{array} \end{equation*}

The order \(dx\ dz\ dy\text{:}\)

The \(x\)-bounds are the same as the order above. We now consider the triangle in Figure 13.6.15(a) and describe it with the order \(dz\ dy\text{:}\) \(0\leq z\leq 2-y/3\) and \(0\leq y\leq 6\text{.}\) Thus the volume is given by: \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq x\leq 3-y/2-3z/2\\ 0\leq z\leq 2-y/3\\ 0\leq y\leq 6 \end{array} \amp \ds\Rightarrow \int_0^6\int_0^{2-y/3}\int_0^{3-y/2-3z/2}\ dx\ dz\ dy. \end{array} \end{equation*}

The order \(dy\ dz\ dx\text{:}\)

We now need to determine the \(y\)-surfaces that determine our region. Approaching the space region from “behind” and moving in the direction of increasing \(y\text{,}\) we first enter the region at \(y=0\text{,}\) and exit along the plane \(z= 2-y/3-2x/3\text{.}\) Solving for \(y\text{,}\) this plane has equation \(y = 6-2x-3z\text{.}\) Thus \(y\) has bounds \(0\leq y\leq 6-2x-3z\text{.}\)

Now collapse the region onto the \(x\)-\(z\) plane, as shown in Figure 13.6.15(b). The curves bounding this triangle are \(z=0\) and \(z=2-2x/3\text{;}\) \(x\) is bounded by the points \(x=0\) to \(x=3\text{.}\) Thus the triple integral giving volume is: \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq y\leq 6-2x-3z\\ 0\leq z\leq 2-2x/3\\ 0\leq x\leq 3 \end{array} \amp \ds\Rightarrow \int_0^3\int_0^{2-2x/3}\int_0^{6-2x-3z}\ dy\ dz\ dx. \end{array} \end{equation*}

The order \(dy\ dx\ dz\text{:}\)

The \(y\)-bounds are the same as in the order above. We now determine the bounds of the triangle in Figure 13.6.15(b) using the order \(dy\ dx\ dz\text{.}\) \(x\) is bounded by \(x=0\) and \(x=3-3z/2\text{;}\) \(z\) is bounded between \(z=0\) and \(z=2\text{.}\) This leads to the triple integral: \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq y\leq 6-2x-3z\\ 0\leq x\leq 3-3z/2\\ 0\leq z\leq 2 \end{array} \amp \ds\Rightarrow \int_0^2\int_0^{3-3z/2}\int_0^{6-2x-3z}\ dy\ dx\ dz. \end{array} \end{equation*}

This problem was long, but hopefully useful, demonstrating how to determine bounds with every order of integration to describe the region \(D\text{.}\) In practice, we only need 1, but being able to do them all gives us flexibility to choose the order that suits us best.

The two surfaces are \(z=3-x^2-y^2\) and \(z=2y\text{.}\) To find where they intersect, it is natural to set them equal to each other: \(3-x^2-y^2=2y\text{.}\) This is an implicit function of \(x\) and \(y\) that gives all points \((x,y)\) in the \(x\)-\(y\) plane where the \(z\) values of the two surfaces are equal.

We can rewrite this implicit function by completing the square: \begin{equation*} 3-x^2-y^2=2y \Rightarrow y^2+2y+x^2=3 \Rightarrow (y+1)^2+x^2=4. \end{equation*}

Thus in the \(x\)-\(y\) plane the projection of the intersection is a circle with radius 2, centered at \((0,-1)\text{.}\)

To project onto the \(x\)-\(z\) plane, we do a similar procedure: find the \(x\) and \(z\) values where the \(y\) values on the surface are the same. We start by solving the equation of each surface for \(y\text{.}\) In this particular case, it works well to actually solve for \(y^2\text{:}\)

\(z=3-x^2-y^2 \Rightarrow y^2=3-x^2-z\)

\(z=2y \Rightarrow y^2=z^2/4\text{.}\)

Thus we have (after again completing the square): \begin{equation*} 3-x^2-z = z^2/4 \Rightarrow \frac{(z+2)^2}{16}+\frac{x^2}4=1, \end{equation*} and ellipse centered at \((0,-2)\) in the \(x\)-\(z\) plane with a major axis of length 8 and a minor axis of length 4.

Finally, to project the curve of intersection into the \(y\)-\(z\) plane, we solve equation for \(x\text{.}\) Since \(z=2y\) is a cylinder that lacks the variable \(x\text{,}\) it becomes our equation of the projection in the \(y\)-\(z\) plane.

All three projections are shown in Figure 13.6.19(b).

The order \(dz\ dy\ dx\text{:}\)

The region \(D\) is bounded below by the plane \(z=0\) and above by the plane \(z=-y\text{.}\) The cylinder \(x^2+y^2=1\) does not offer any bounds in the \(z\)-direction, as that surface is parallel to the \(z\)-axis. Thus \(0\leq z\leq -y\text{.}\)

Collapsing the region into the \(x\)-\(y\) plane, we get part of the circle with equation \(x^2+y^2=1\) as shown in Figure 13.6.23(b). As a function of \(x\text{,}\) this half circle has equation \(y=-\sqrt{1-x^2}\text{.}\) Thus \(y\) is bounded below by \(-\sqrt{1-x^2}\) and above by \(y=0\text{:}\) \(-\sqrt{1-x^2}\leq y\leq 0\text{.}\) The \(x\) bounds of the half circle are \(-1\leq x\leq 1\text{.}\) All together, the bounds of integration and triple integral are as follows: \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq z\leq -y\\ -\sqrt{1-x^2}\leq y\leq 0\\ -1\leq x\leq 1 \end{array} \amp \ds\Rightarrow \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\int_0^{-y}\ dz\ dy\ dx. \end{array} \end{equation*}

We evaluate this triple integral: \begin{align*} \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\int_0^{-y}\ dz\ dy\ dx \amp = \int_{-1}^1\int_{-\sqrt{1-x^2}}^{0}\big(-y\big)\ dy\ dx\\ \amp =\int_{-1}^1\big(-\frac12y^2\big)\Big|_{-\sqrt{1-x^2}}^{0}\ dx\\ \amp = \int_{-1}^1 \frac12\big(1-x^2\big)\ dx\\ \amp = \left.\left(\frac12\left(x-\frac13x^3\right)\right)\right|_{-1}^1\\ \amp = \frac23\text{ units } ^3. \end{align*}

With the order \(dy\ dx\ dz\text{:}\)

The region is bounded “below” in the \(y\)-direction by the surface \(x^2+y^2=1 \Rightarrow y=-\sqrt{1-x^2}\) and “above” by the surface \(y=-z\text{.}\) Thus the \(y\) bounds are \(-\sqrt{1-x^2}\leq y\leq -z\text{.}\)

Collapsing the region onto the \(x\)-\(z\) plane gives the region shown in Figure 13.6.26(a); this half circle has equation \(x^2+z^2=1\text{.}\) (We find this curve by solving each surface for \(y^2\text{,}\) then setting them equal to each other. We have \(y^2=1-x^2\) and \(y=-z\Rightarrow y^2=z^2\text{.}\) Thus \(x^2+z^2=1\text{.}\)) It is bounded below by \(x=-\sqrt{1-z^2}\) and above by \(x=\sqrt{1-z^2}\text{,}\) where \(z\) is bounded by \(0\leq z\leq 1\text{.}\) All together, we have: \begin{equation*} \begin{array}{cc} \begin{array}{c} -\sqrt{1-x^2}\leq y\leq -z\\ -\sqrt{1-z^2}\leq x\leq \sqrt{1-z^2}\\ 0\leq z\leq 1 \end{array} \amp \ds\Rightarrow \int_{0}^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{-\sqrt{1-x^2}}^{-z}\ dy\ dx\ dz. \end{array} \end{equation*}

With the order \(dx\ dz\ dy\text{:}\)

\(D\) is bounded below by the surface \(x=-\sqrt{1-y^2}\) and above by \(\sqrt{1-y^2}\text{.}\) We then collapse the region onto the \(y\)-\(z\) plane and get the triangle shown in Figure 13.6.26(b). (The hypotenuse is the line \(z=-y\text{,}\) just as the plane.) Thus \(z\) is bounded by \(0\leq z\leq -y\) and \(y\) is bounded by \(-1\leq y\leq 0\text{.}\) This gives: \begin{equation*} \begin{array}{cc} \begin{array}{c} -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}\\ 0\leq z\leq -y\\ -1\leq y\leq 0 \end{array} \amp \ds\Rightarrow \int_{-1}^0\int_{0}^{-y}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\ dx\ dz\ dy. \end{array} \end{equation*}

Following the bounds–determining strategy of “surface to surface, curve to curve, and point to point,” we can see that the most difficult orders of integration are the two in which we integrate with respect to \(z\) first, for there are two “upper” surfaces that bound \(D\) in the \(z\)-direction. So we start by noting that we have \begin{equation*} 0\leq z\leq 1-\frac12x \text{ and } 0\leq z\leq 1-\frac14y. \end{equation*}

We now collapse the region \(D\) onto the \(x\)-\(y\) axis, as shown in Figure 13.6.31(b). The boundary of \(D\text{,}\) the line from \((0,0,1)\) to \((2,4,0)\text{,}\) is shown in part (b) of the figure as a dashed line; it has equation \(y=2x\text{.}\) (We can recognize this in two ways: one, in collapsing the line from \((0,0,1)\) to \((2,4,0)\) onto the \(x\)-\(y\) plane, we simply ignore the \(z\)-values, meaning the line now goes from \((0,0)\) to \((2,4)\text{.}\) Secondly, the two surfaces meet where \(z=1-x/2\) is equal to \(z=1-y/4\text{:}\) thus \(1-x/2=1-y/4 \Rightarrow y=2x\text{.}\))

We use the second property of Theorem 13.6.29 to state that \begin{equation*} \iiint_D \ dV = \iiint_{D_1}\ dV + \iiint_{D_2}\ dV, \end{equation*} where \(D_1\) and \(D_2\) are the space regions above the plane regions \(R_1\) and \(R_2\text{,}\) respectively. Thus we can say \begin{equation*} \iiint_D\ dV = \iint_{R_1}\left(\int_0^{1-x/2}\ dz\right)dA + \iint_{R_2}\left(\int_0^{1-y/4}\ dz\right)dA. \end{equation*}

All that is left is to determine bounds of \(R_1\) and \(R_2\text{,}\) depending on whether we are integrating with order \(dx\ dy\) or \(dy\ dx\text{.}\) We give the final integrals here, leaving it to the reader to confirm these results.

\(dz\ dy\ dx\text{:}\) \begin{equation*} \begin{array}{ccccc} \amp \amp \begin{array}{c} 0\leq z\leq 1-x/2\\ 0\leq y\leq 2x\\ 0\leq x\leq 2 \end{array} \amp \amp \begin{array}{c} 0\leq z\leq 1-y/4\\ 2x\leq y\leq 4\\ 0\leq x\leq 2 \end{array} \\ \\ \ds \iiint_D\ dV \amp =\amp \ds \int_0^2\int_0^{2x}\int_0^{1-x/2}\ dz\ dy\ dx \amp +\amp \ds \int_0^2\int_{2x}^4\int_0^{1-y/4}\ dz\ dy\ dx \end{array} \end{equation*}

\(dz\ dx\ dy\text{:}\) \begin{equation*} \begin{array}{ccccc} \amp \amp \begin{array}{c} 0\leq z\leq 1-x/2\\ y/2\leq x\leq 2\\ 0\leq y\leq 4 \end{array} \amp \amp \begin{array}{c} 0\leq z\leq 1-y/4\\ 0\leq x\leq y/2\\ 0\leq y\leq 4 \end{array} \\ \\ \ds \iiint_D\ dV \amp =\amp \ds \int_0^4\int_{y/2}^{2}\int_0^{1-x/2}\ dz\ dx\ dy \amp +\amp \ds \int_0^4\int_{0}^{y/2}\int_0^{1-y/4}\ dz\ dx\ dy \end{array} \end{equation*}

The remaining four orders of integration do not require a sum of triple integrals. In Figure 13.6.34 we show \(D\) collapsed onto the other two coordinate planes. Using these graphs, we give the final orders of integration here, again leaving it to the reader to confirm these results.

\(dy\ dx\ dz\text{:}\) \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq y\leq 4-4z\\ 0\leq x\leq 2-2z\\ 0\leq z\leq 1 \end{array} \amp \ds \Rightarrow \int_0^1\int_{0}^{2-2z}\int_0^{4-4z}\ dy\ dx\ dz \end{array} \end{equation*}

\(dy\ dz\ dx\text{:}\) \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq y\leq 4-4z\\ 0\leq z\leq 1-x/2\\ 0\leq x\leq 2 \end{array} \amp \ds \Rightarrow \int_0^2\int_{0}^{1-x/2}\int_0^{4-4z}\ dy\ dx\ dz \end{array} \end{equation*}

\(dx\ dy\ dz\text{:}\) \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq x\leq 2-2z\\ 0\leq y\leq 4-4z\\ 0\leq z\leq 1 \end{array} \amp \ds \Rightarrow \int_0^1\int_{0}^{4-4z}\int_0^{2-2z}\ dx\ dy\ dz \end{array} \end{equation*}

\(dx\ dz\ dy\text{:}\) \begin{equation*} \begin{array}{cc} \begin{array}{c} 0\leq x\leq 2-2z\\ 0\leq z\leq 1-y/4\\ 0\leq y\leq 4 \end{array} \amp \ds \Rightarrow \int_0^4\int_{0}^{1-y/4}\int_0^{2-2z}\ dx\ dz\ dy \end{array} \end{equation*}

The main point of this example is this: integrating with respect to \(z\) first is rather straightforward; integrating with respect to \(x\) first is not.

The order \(dz\ dy\ dx\text{:}\)

The bounds on \(z\) are clearly \(2x^2+2\leq z\leq 6-2x^2-y^2\text{.}\) Collapsing \(D\) onto the \(x\)-\(y\) plane gives the ellipse shown in Figure 13.6.38(c). The equation of this ellipse is found by setting the two surfaces equal to each other: \begin{equation*} 2x^2+2 = 6-2x^2-y^2 \Rightarrow 4x^2+y^2=4 \Rightarrow x^2+\frac{y^2}4=1. \end{equation*}

\hskip2 We can describe this ellipse with the bounds \begin{equation*} -\sqrt{4-4x^2} \leq y\leq \sqrt{4-4x^2} \text{ and } -1\leq x\leq 1. \end{equation*}

Thus we find volume as \begin{equation*} \begin{array}{cc} \begin{array}{c} 2x^2+2\leq z\leq 6-2x^2-y^2\\ -\sqrt{4-4x^2}\leq y\leq \sqrt{4-4x^2}\\ -1\leq x\leq 1 \end{array} \amp \ds \Rightarrow \int_{-1}^1\int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}}\int_{2x^2+2}^{6-2x^2-y^2}\ dz\ dy\ dx \end{array} . \end{equation*}

The order \(dy\ dz\ dx\text{:}\)

Integrating with respect to \(y\) is not too difficult. Since the surface \(z=2x^2+2\) is a cylinder whose directrix is the \(y\)-axis, it does not create a border for \(y\text{.}\) The paraboloid \(z=6-2x^2-y^2\) does; solving for \(y\text{,}\) we get the bounds \begin{equation*} -\sqrt{6-2x^2-z}\leq y\leq \sqrt{6-2x^2-z}. \end{equation*}

Collapsing \(D\) onto the \(x\)-\(z\) axes gives the region shown in Figure 13.6.42(a); the lower curve is from the cylinder, with equation \(z=2x^2+2\text{.}\) The upper curve is from the paraboloid; with \(y=0\text{,}\) the curve is \(z=6-2x^2\text{.}\) Thus bounds on \(z\) are \(2x^2+2\leq z\leq 6-2x^2\text{;}\) the bounds on \(x\) are \(-1\leq x\leq 1\text{.}\) Thus we have: \begin{equation*} \begin{array}{cc} \begin{array}{c} -\sqrt{6-2x^2-z}\leq y\leq \sqrt{6-2x^2-z}\\ 2x^2+2\leq z\leq 6-2x^2\\ -1\leq x\leq 1 \end{array} \amp \ds \Rightarrow \int_{-1}^1\int_{2x^2+2}^{6-2x^2}\int_{-\sqrt{6-2x^2-z}}^{\sqrt{6-2x^2-z}}\ dy\ dz\ dx. \end{array} \end{equation*}

The order \(dx\ dz\ dy\text{:}\)

This order takes more effort as \(D\) must be split into two subregions. The two surfaces create two sets of upper/lower bounds in terms of \(x\text{;}\) the cylinder creates bounds \begin{equation*} -\sqrt{z/2-1}\leq x\leq \sqrt{z/2-1} \end{equation*} for region \(D_1\) and the paraboloid creates bounds \begin{equation*} -\sqrt{3-y^2/2-z^2/2}\leq x\leq \sqrt{3-y^2/2-z^2/2} \end{equation*} for region \(D_2\text{.}\)

Collapsing \(D\) onto the \(y\)-\(z\) axes gives the regions shown in Figure 13.6.42(b). We find the equation of the curve \(z=4-y^2/2\) by noting that the equation of the ellipse seen in Figure 13.6.38(c) has equation \begin{equation*} x^2+y^2/4=1 \Rightarrow x = \sqrt{1-y^2/4}. \end{equation*}

Substitute this expression for \(x\) in either surface equation, \(z=6-2x^2-y^2\) or \(z=2x^2+2\text{.}\) In both cases, we find \begin{equation*} z=4-\frac12y^2. \end{equation*}

Region \(R_1\text{,}\) corresponding to \(D_1\text{,}\) has bounds \begin{equation*} 2\leq z\leq 4-y^2/2, -2\leq y\leq 2 \end{equation*} and region \(R_2\text{,}\) corresponding to \(D_2\text{,}\) has bounds \begin{equation*} 4-y^2/2\leq z\leq 6-y^2, -2\leq y\leq 2. \end{equation*}

Thus the volume of \(D\) is given by: \begin{equation*} \int_{-2}^2\int_2^{4-y^2/2}\int_{-\sqrt{z/2-1}}^{\sqrt{z/2-1}}\ dx\ dz\ dy \ +\ \int_{-2}^2\int_{4-y^2/2}^{6-y^2}\int_{-\sqrt{3-y^2/2-z^2/2}}^{\sqrt{3-y^2/2-z^2/2}}\ dx\ dz\ dy. \end{equation*}

We evaluate this integral according to Definition 13.6.45.

\(\ds \int_0^1\int_{x^2}^x\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\ dz\ dy\ dx\) \begin{align*} \amp = \int_0^1\int_{x^2}^x\left(\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\ dz\right)\ dy\ dx\\ \amp = \int_0^1\int_{x^2}^x\left(\big(xyz+ xz^2\big)\Big|_{x^2-y}^{2x+3y}\right)\ dy\ dx\\ \amp = \int_0^1\int_{x^2}^x\Bigg(xy(2x+3y)+x(2x+3y)^2-\Big(xy(x^2-y)+x(x^2-y)^2\Big)\Bigg)\ dy\ dx\\ \amp =\int_0^1\int_{x^2}^x\Big(-x^5+x^3y+4x^3+14x^2y+12xy^2\Big)\ dy\ dx.\\ \end{align*} We continue as we have in the past, showing fewer steps. \begin{align*} \amp = \int_0^1\Bigg(-\frac72x^7-8x^6-\frac72x^5+15x^4\Bigg)\ dx\\ \amp = \frac{281}{336}\approx 0.836. \end{align*}