Okay, this first one is set up to just use the formula above so let’s do that.

\[\begin{align*}\iint\limits_{D}{{{{\bf{e}}^{\frac{x}{y}}}\,dA}} & = \int_{{\,1}}^{{\,2}}{{\int_{{\,y}}^{{{y^3}}}{{{{\bf{e}}^{\frac{x}{y}}}\,dx}}\,dy}} = \int_{{\,1}}^{{\,2}}{{\left. {y\,{{\bf{e}}^{\frac{x}{y}}}} \right|_y^{{y^3}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{y\,{{\bf{e}}^{{y^2}}} - y{{\bf{e}}^1}\,dy}}\\ & = \left. {\left( {\frac{1}{2}{{\bf{e}}^{{y^2}}} - \frac{1}{2}{y^2}{{\bf{e}}^1}} \right)} \right|_1^2 = \frac{1}{2}{{\bf{e}}^4} - 2{{\bf{e}}^1}\end{align*}\]

In this case we need to determine the two inequalities for \(x\) and \(y\) that we need to do the integral. The best way to do this is the graph the two curves. Here is a sketch.

So, from the sketch we can see that that two inequalities are,

\[0 \le x \le 1\hspace{0.25in}{x^3} \le y \le \sqrt x \]

We can now do the integral,

\[\begin{align*}\iint\limits_{D}{{4xy - {y^3}\,dA}} & = \int_{{\,0}}^{{\,1}}{{\int_{{\,{x^3}}}^{{\,\sqrt x }}{{4xy - {y^3}\,dy}}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\left. {\left( {2x{y^2} - \frac{1}{4}{y^4}} \right)} \right|_{{x^3}}^{\sqrt x }\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\frac{7}{4}{x^2} - 2{x^7} + \frac{1}{4}{x^{12}}\,dx}}\\ & = \left. {\left( {\frac{7}{{12}}{x^3} - \frac{1}{4}{x^8} + \frac{1}{{52}}{x^{13}}} \right)} \right|_0^1 = \frac{{55}}{{156}}\end{align*}\]

We got even less information about the region this time. Let’s start this off by sketching the triangle.

Since we have two points on each edge it is easy to get the equations for each edge and so we’ll leave it to you to verify the equations.

Now, there are two ways to describe this region. If we use functions of \(x\), as shown in the image we will have to break the region up into two different pieces since the lower function is different depending upon the value of \(x\). In this case the region would be given by \(D = {D_1} \cup {D_2}\) where,

\[\begin{align*}{D_1} & = \left\{ {\left( {x,y} \right)|0 \le x \le 1,\,\,\, - 2x + 3 \le y \le 3} \right\}\\ {D_2} & = \left\{ {\left( {x,y} \right)|1 \le x \le 5,\,\,\,\frac{1}{2}x + \frac{1}{2} \le y \le 3} \right\}\end{align*}\]

Note the \( \cup \) is the “union” symbol and just means that \(D\) is the region we get by combing the two regions. If we do this then we’ll need to do two separate integrals, one for each of the regions.

To avoid this we could turn things around and solve the two equations for \(x\) to get,

\[\begin{align*}y & = - 2x + 3\hspace{0.5in} \Rightarrow \hspace{0.5in}x = - \frac{1}{2}y + \frac{3}{2}\\ y & = \frac{1}{2}x + \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}x = 2y - 1\end{align*}\]

If we do this we can notice that the same function is always on the right and the same function is always on the left and so the region is,

\[D = \left\{ {\left( {x,y} \right)|\,\, - \frac{1}{2}y + \frac{3}{2} \le x \le 2y - 1,\,\,\,1 \le y \le 3} \right\}\]

Writing the region in this form means doing a single integral instead of the two integrals we’d have to do otherwise.

Either way should give the same answer and so we can get an example in the notes of splitting a region up let’s do both integrals.

Solution 1

\[\begin{align*}\iint\limits_{D}{{6{x^2} - 40y\,dA}} & = \iint\limits_{{{D_1}}}{{6{x^2} - 40y\,dA}} + \iint\limits_{{{D_2}}}{{6{x^2} - 40y\,dA}}\\ & = \int_{{\,0}}^{{\,1}}{{\int_{{\, - 2x + 3}}^{{\,3}}{{6{x^2} - 40y\,dy}}\,dx}} + \int_{{\,1}}^{{\,5}}{{\int_{{\,\frac{1}{2}x + \frac{1}{2}}}^{{\,3}}{{6{x^2} - 40y\,dy}}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\left. {\left( {6{x^2}y - 20{y^2}} \right)} \right|_{ - 2x + 3}^3\,dx}} + \int_{{\,1}}^{{\,5}}{{\left. {\left( {6{x^2}y - 20{y^2}} \right)} \right|_{\frac{1}{2}x + \frac{1}{2}}^3\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{12{x^3} - 180 + 20{{\left( {3 - 2x} \right)}^2}\,dx}} + \int_{{\,1}}^{{\,5}}{{ - 3{x^3} + 15{x^2} - 180 + 20{{\left( {\frac{1}{2}x + \frac{1}{2}} \right)}^2}\,dx}}\\ & = \left. {\left( {3{x^4} - 180x - \frac{{10}}{3}{{\left( {3 - 2x} \right)}^3}} \right)} \right|_0^1 + \left. {\left( { - \frac{3}{4}{x^4} + 5{x^3} - 180x + \frac{{40}}{3}{{\left( {\frac{1}{2}x + \frac{1}{2}} \right)}^3}} \right)} \right|_1^5\\ & = - \frac{{935}}{3}\end{align*}\]

That was a lot of work. Notice however, that after we did the first substitution that we didn’t multiply everything out. The two quadratic terms can be easily integrated with a basic Calc I substitution and so we didn’t bother to multiply them out. We’ll do that on occasion to make some of these integrals a little easier.

Solution 2
This solution will be a lot less work since we are only going to do a single integral.

\[\begin{align*}\iint\limits_{D}{{6{x^2} - 40y\,dA}} &= \int_{1}^{3}{{\int_{{ - \frac{1}{2}y + \frac{3}{2}}}^{{2y - 1}}{{6{x^2} - 40y\,dx}}\,dy}}\\ & = \int_{1}^{3}{{\left. {\left( {2{x^3} - 40xy} \right)} \right|_{ - \frac{1}{2}y + \frac{3}{2}}^{2y - 1}\,dy}}\\ & = \int_{1}^{3}{{100y - 100{y^2} + 2{{\left( {2y - 1} \right)}^3} - 2{{\left( { - \frac{1}{2}y + \frac{3}{2}} \right)}^3}\,dy}}\\ & = \left. {\left( {50{y^2} - \frac{{100}}{3}{y^3} + \frac{1}{4}{{\left( {2y - 1} \right)}^4} + {{\left( { - \frac{1}{2}y + \frac{3}{2}} \right)}^4}} \right)} \right|_1^3\\ & = - \frac{{935}}{3}\end{align*}\]

So, the numbers were a little messier, but other than that there was much less work for the same result. Also notice that again we didn’t cube out the two terms as they are easier to deal with using a Calc I substitution.

First, notice that if we try to integrate with respect to \(y\) we can’t do the integral because we would need a \({y^2}\) in front of the exponential in order to do the \(y\) integration. We are going to hope that if we reverse the order of integration we will get an integral that we can do.

Now, when we say that we’re going to reverse the order of integration this means that we want to integrate with respect to \(x\) first and then \(y\). Note as well that we can’t just interchange the integrals, keeping the original limits, and be done with it. This would not fix our original problem and in order to integrate with respect to \(x\) we can’t have \(x\)’s in the limits of the integrals. Even if we ignored that the answer would not be a constant as it should be.

So, let’s see how we reverse the order of integration. The best way to reverse the order of integration is to first sketch the region given by the original limits of integration. From the integral we see that the inequalities that define this region are,

\[\begin{array}{c}0 \le x \le 3\\ {x^2} \le y \le 9\end{array}\]

These inequalities tell us that we want the region with \(y = {x^2}\) on the lower boundary and \(y = 9\) on the upper boundary that lies between \(x = 0\) and \(x = 3\). Here is a sketch of that region.

Since we want to integrate with respect to \(x\) first we will need to determine limits of \(x\) (probably in terms of \(y\)) and then get the limits on the \(y\)’s. Here they are for this region.

\[\begin{array}{c}0 \le x \le \sqrt y \\ 0 \le y \le 9\end{array}\]

Any horizontal line drawn in this region will start at \(x = 0\) and end at \(x = \sqrt y \) and so these are the limits on the \(x\)’s and the range of \(y\)’s for the regions is 0 to 9.

The integral, with the order reversed, is now,

\[\int_{{\,0}}^{{\,3}}{{\int_{{\,{x^2}}}^{{\,9}}{{{x^3}{{\bf{e}}^{{y^3}}}\,dy}}\,dx}} = \int_{{\,0}}^{{\,9}}{{\int_{{\,0}}^{{\,\sqrt y }}{{{x^3}{{\bf{e}}^{{y^3}}}\,dx}}\,dy}}\]

and notice that we can do the first integration with this order. We’ll also hope that this will give us a second integral that we can do. Here is the work for this integral.

\[\begin{align*}\int_{{\,0}}^{{\,3}}{{\int_{{\,{x^2}}}^{{\,9}}{{{x^3}{{\bf{e}}^{{y^3}}}\,dy}}\,dx}} & = \int_{{\,0}}^{{\,9}}{{\int_{{\,0}}^{{\,\sqrt y }}{{{x^3}{{\bf{e}}^{{y^3}}}\,dx}}\,dy}}\\ & = \int_{{\,0}}^{{\,9}}{{\left. {\frac{1}{4}{x^4}{{\bf{e}}^{{y^3}}}} \right|_0^{\sqrt y }\,dy}}\\ & = \int_{{\,0}}^{{\,9}}{{\frac{1}{4}{y^2}{{\bf{e}}^{{y^3}}}\,dy}}\\ & = \left. {\frac{1}{{12}}{{\bf{e}}^{{y^3}}}} \right|_0^9\\ & = \frac{1}{{12}}\left( {{{\bf{e}}^{729}} - 1} \right)\end{align*}\]

So, as we hoped, we were able to do the integral once we interchanged the order of integration.

As with the first integral we cannot do this integral by integrating with respect to \(x\) first so we’ll hope that by reversing the order of integration we will get something that we can integrate. Here are the limits for the variables that we get from this integral.

\[\begin{array}{c}\sqrt[3]{y} \le x \le 2\\ 0 \le y \le 8\end{array}\]

and here is a sketch of this region.

So, if we reverse the order of integration we get the following limits.

\[\begin{array}{c}0 \le x \le 2\\ 0 \le y \le {x^3}\end{array}\]

The integral is then,

\[\begin{align*}\int_{{\,0}}^{{\,8}}{{\int_{{\,\sqrt[3]{y}}}^{{\,2}}{{\sqrt {{x^4} + 1} \,dx}}\,dy}} & = \int_{{\,0}}^{{\,2}}{{\int_{{\,0}}^{{\,{x^3}}}{{\sqrt {{x^4} + 1} \,dy}}\,dx}}\\ & = \int_{{\,0}}^{{\,2}}{{\left. {y\sqrt {{x^4} + 1} } \right|_0^{{x^3}}\,dx}}\\ & = \int_{{\,0}}^{{\,2}}{{{x^3}\sqrt {{x^4} + 1} \,dx}} = \frac{1}{6}\left( {{{17}^{\frac{3}{2}}} - 1} \right)\end{align*}\]

Here is the graph of the surface and we’ve tried to show the region in the \(xy\)-plane below the surface.

Here is a sketch of the region in the \(xy\)-plane by itself.

By setting the two bounding equations equal we can see that they will intersect at \(x = 2\) and \(x = - 2\). So, the inequalities that will define the region \(D\) in the \(xy\)-plane are,

\[\begin{array}{c} - 2 \le x \le 2\\ {x^2} \le y \le 8 - {x^2}\end{array}\]

The volume is then given by,

\[\begin{align*}V &= \iint\limits_{D}{{16xy + 200\,dA}}\\ & = \int_{{ - 2}}^{{\,2}}{{\int_{{{x^2}}}^{{8 - {x^2}}}{{16xy + 200\,dy}}\,dx}}\\ & = \int_{{ - 2}}^{{\,2}}{{\left. {\left( {8x{y^2} + 200y} \right)} \right|_{{x^2}}^{8 - {x^2}}\,dx}}\\ & = \int_{{ - 2}}^{{\,2}}{{ - 128{x^3} - 400{x^2} + 512x + 1600\,dx}}\\ & = \left. {\left( { - 32{x^4} - \frac{{400}}{3}{x^3} + 256{x^2} + 1600x} \right)} \right|_{ - 2}^2 = \frac{{12800}}{3}\end{align*}\]

This example is a little different from the previous one. Here the region \(D\) is not explicitly given so we’re going to have to find it. First, notice that the last two planes are really telling us that we won’t go past the \(xy\)-plane and the \(yz\)-plane when we reach them.

The first plane, \(4x + 2y + z = 10\), is the top of the volume and so we are really looking for the volume under,

\[z = 10 - 4x - 2y\]

and above the region \(D\) in the \(xy\)-plane. The second plane, \(y = 3x\) (yes that is a plane), gives one of the sides of the volume as shown below.

The region \(D\) will be the region in the \(xy\)-plane (i.e. \(z = 0\)) that is bounded by \(y = 3x\), \(x = 0\), and the line where \(z + 4x + 2y = 10\) intersects the \(xy\)-plane. We can determine where \(z + 4x + 2y = 10\) intersects the \(xy\)-plane by plugging \(z = 0\) into it.

\[0 + 4x + 2y = 10\hspace{0.25in} \Rightarrow \hspace{0.25in}2x + y = 5\hspace{0.25in} \Rightarrow \hspace{0.25in}y = - 2x + 5\]

So, here is a sketch the region \(D\).

The region \(D\) is really where this solid will sit on the \(xy\)-plane and here are the inequalities that define the region.

\[\begin{array}{c}0 \le x \le 1\\ 3x \le y \le - 2x + 5\end{array}\]

Here is the volume of this solid.

\[\begin{align*}V & = \iint\limits_{D}{{10 - 4x - 2y\,dA}}\\ & = \int_{0}^{{\,1}}{{\int_{{3x}}^{{ - 2x + 5}}{{10 - 4x - 2y\,dy}}\,dx}}\\ & = \int_{0}^{{\,1}}{{\left. {\left( {10y - 4xy - {y^2}} \right)} \right|_{3x}^{ - 2x + 5}\,dx}}\\ & = \int_{0}^{{\,1}}{{25{x^2} - 50x + 25\,dx}}\\ & = \left. {\left( {\frac{{25}}{3}{x^3} - 25{x^2} + 25x} \right)} \right|_0^1 = \frac{{25}}{3}\end{align*}\]