How To Find The Bounds Of A Triple Integral

[Paula Shuffleburg]

Althoughwe define triple integrals using a Riemann sum, we usually evaluate triple integrals by turning them into iterated integrals involving three single integrals. One tricky part of triple integrals is describing the three-dimensional regions of integration and the resulting bounds on the iterated integrals. Forming double iterated integrals is easier because one can draw the domain and label all the edges and corners, which makes determining the bounds more tractable. Three-dimensional regions are much more difficult to visualize or draw, which can make the prospect of determining integration bounds a formidable task.

In the shadow method, you imagine there is a light source, such as the sun, positioned far away along one of the coordinate axis (such as the positive $z$-axis). We'll think of this sun as being straight up in the sky and think of the chosen coordinate axis as though it were vertical.

Putting all the limits of integration together and inserting the integrand $f(x,y,z)=xy$, we calculate the integral:\beginalign*\iiint_\dlvf(x,y,z)dV &=\int_0^1 \int_0^y \int_x-y^2+x-y xy\, dz\,dx\,dy\\&= \int_0^1 \int_0^y xyz \bigg_z=x-y^z=2+x-y dx\,dy\\&= \int_0^1 \int_0^y 2xy dx\,dy\\&= \int_0^1 x^2y \bigg_x=0^x=ydy\\&= \int_0^1 y^3 dy = \fracy^44\bigg_0^1 = \frac14.\endalign*Example 2Let $\rho(x,y,z)$ be the charge density at the point $(x,y,z)$ inside an object $\dlv$ bounded by the elliptic paraboloids $y=5-4 x^2-z^2$ and $y=x^2+z^2/4$. Set up the integral giving the total charge inside $\dlv$.

Region bounded by paraboloids demonstrating the shadow method. The region $\dlv$ is bounded by the elliptic paraboloids $y=5-4 x^2-z^2$ and $y=x^2+z^2/4$. If the sun is viewed to be in the direction of the positive $y$-axis, the shadow of $\dlv$ in a plane perpendicular to the $y$-axis is the region inside the ellipse $x^2+z^2/4=1$.

The remaining task is to determine the shadow of $\dlv$, illustrated by the ellipse, above. The widest part of $\dlv$ is where the two paraboloids intersect, i.e., where$$x^2+\fracz^24 = 5-4 x^2-z^2,$$which can be written more simply as$$x^2+\fracz^24 = 1.$$This boundary of the shadow is an ellipse with a semi-major axis of 2 and a semi-minor axis of 1. The shadow itself is the region$$x^2+\fracz^24 \le 1,$$which includes the interior of the ellipse, shown below.

Since inside the ellipse, $x^2 \le 1- z^2/4$, the range of $x$ for each value of $z$ is $$ -\sqrt1-z^2/4 \le x \le \sqrt1-z^2/4.$$The maximum range of $z$ is $-2 \le z \le 2$, so the integration limits for the shadow are\begingather* \iint_\textshadow \cdots dx\,dz = \int_-2^2 \int_-\sqrt1-z^2/4^\sqrt1-z^2/4 \cdots dx\,dz.\endgather*

Putting the top and bottom limits together with the shadow, we conclude that the total charge inside $\dlv$ is the integral\beginalign* \texttotal charge &= \iiint_\dlv \rho(x,y,z) \, dV\\ &=\int_-2^2 \int_-\sqrt1-z^2/4^\sqrt1-z^2/4\int_x^2+z^2/4^5-4 x^2-z^2\rho(x,y,z)dy\,dx\,dz.\endalign*

The shadow method for determining triple integral bounds by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. For permissions beyond the scope of this license, please contact us.

My problem is with finding the bounds for the integral. I set the $z$s equal to each other and found the intersection is $x^2 + \frac34(y-\frac49)^2 = \frac139$ (although I may have made a mistake here.)

Now I am not sure what to do. How can I find the bounds for $x$ and $y$? Also, I think I should do a change of variable so that the plane lies flat across the cone instead of slanted, to make the upper bound for $z$ constant. Is this a good idea? How could I do it? I am really lost with this problem.

The intersection is indeed an ellipse, but you seem to have made some algebra mistakes:\beginalign*x^2 + y^2 &= (\tfracy + 22)^2 \\x^2 + y^2 &= \tfrac14y^2 + y + 1 \\x^2 + \tfrac34(y^2 - \tfrac43y) &= 1 \\x^2 + \tfrac34(y^2 - \tfrac43y + \tfrac49 - \tfrac49) &= 1 \\x^2 + \tfrac34(y^2 - \tfrac43y + \tfrac49) - \tfrac13 &= 1 \\x^2 + \tfrac34(y - \tfrac23)^2 &= \tfrac43 \\\endalign*Using the shadow method, notice that the above ellipse is the desired shadow that we want to integrate our $x$ and $y$ over. Solving for $x$ (for example) in the above ellipse yields:$$x = \pm \sqrt\tfrac43 - \tfrac34(y - \tfrac23)^2$$A quick plot tells us that the min/max values of $y$ occur at the $y$-intercepts, where $y = \tfrac23 \pm \tfrac43$.so our triple integral is:$$V = \int_-2/3^2 \int_-\sqrt\frac43 - \frac34(y - \frac23)^2^\sqrt\frac43 - \frac34(y - \frac23)^2 \int_\sqrtx^2 + y^2^\fracy + 22 \, dz \, dx \, dy$$

I'm able to solve integrals and draw graphs in 2D, but I have a lot of trouble drawing 3D graphs. As such, I'm seeking a method of finding the bounds of integration for triple integrals without having to graph them. It seems reasonable to me that there is such a method, since we also need a way to solve higher dimensional integrals (4D, 5D, 6D, etc), without the ability to graph them.

These equations don't give the points $(x,y,z)$ which belong to the solid, they only give the points on the surfaces which form the boundary of the solid. So we need to figure out how to make them into inequalities that really describe the solid.

\beginequation

\beginaligned

&=\int_c^d \int_a^b \int_j^k f(x, y, z) d z d x d y \\

&=\int_a^b \int_j^k \int_c^d f(x, y, z) d y d z d x \\

&=\int_j^b \int_a^b \int_c^d f(x, y, z) d y d x d z \\

&=\int_c^d \int_j^k \int_a^b f(x, y, z) d x d z d y \\

&=\int_j^k \int_c^d \int_a^b f(x, y, z) d x d y d z

\endaligned

\endequation

\beginequation

\begingathered

\int_0^2 \int_0^3 \int_-1^2\left(6 x^3 y^2 z\right) d z d y d x=\int_0^2 \int_0^3\left(\int_-1^2\left(6 x^3 y^2 z\right) d z\right) d y d x \\

=\int_0^2 \int_0^3\left(\left[3 x^3 y^2 z^2\right]_-1^2\right) d y d x \\

=\int_0^2\left(\int_0^3\left(9 x^3 y^2\right) d y\right) d x \\

=\int_0^2\left(\left[3 x^3 y^3\right]_0^3\right) d x \\

=\int_0^2\left(81 x^3\right) d x \\

=\left(\frac814 x^4\right]_0^2 \\

=324

\endgathered

\endequation

But as we learned from double integrals over non-rectangular regions, the most challenging part of triple integration over general regions is determining the limits (bounds) for each integral.

\beginequation

\begingathered

=\int_0^1 \int_0^y\left(\left[\fracz^22\right]_0^\sqrt1-y^2\right) d x d y \\

\left.=\int_0^1 \int_0^y \int_0^y\left(\frac12\left(1-y^2\right)\right) d x\right) d y \\

=\int_0^1\left(\left[\frac12 x\left(1-y^2\right)\right]_0^y\right) d y \\

=\int_0^1\left(\frac12\left(y-y^3\right)\right) d y \\

=\left(\frac12\left(\fracy^22-\fracy^44\right)\right]_0^1 \\

=\frac18

\endgathered

\endequation

Together in our lesson, we will walk through several examples in detail to ensure that we can evaluate an iterated triple integral to find the volume of a solid. And we will devote our energy to learning how to identify our limits of integration, as they can be prickly at times.

I'm studying for an exam and I struggle with finding the bounds for triple integrals. Specifically, I find it difficult when I can not draw a picture of the surface (which I may not have time to do on my exam anyway).

Also, my professor suggests doing the integral over the vertical variable (z in this case) first, before trying to find the bounds for the other variables (then just doing a double integral). If I haven't found the bounds for the others yet, and I don't know what the region looks like, how do I determine which is the upper bound vs. lower bound for $z$?

The upper and lower bounds for $z$ in the triple integral would be the two functions you started with. So integrating w.r.t. $z$ would just add the step of subtracting the lower bound from the upper bound, as you've already done.

A triple integral is a mathematical tool used to calculate the volume of a three-dimensional object. It is set up by integrating a function over a three-dimensional region, with each integral representing a different dimension.

The limits of integration for a triple integral depend on the shape of the three-dimensional region and the coordinate system being used. They can be determined by visualizing the region and breaking it down into smaller, simpler shapes.

A triple integral in rectangular coordinates is set up using rectangular boundaries, while a triple integral in cylindrical or spherical coordinates is set up using polar or spherical boundaries. This allows for more efficient and accurate integration for certain types of three-dimensional objects.

No, a triple integral can only be used to find the volume of objects that can be represented by a continuous function over a three-dimensional region. For more complex objects, other methods such as surface integration may be necessary.

Triple integrals have many applications in physics, engineering, and other fields. They can be used to calculate the mass of an object with varying density, find the center of mass of a three-dimensional object, and determine fluid flow or electric field strength over a three-dimensional region.

In this section we want do take a look at triple integrals done completely in Cylindrical Coordinates. Recall that cylindrical coordinates are really nothing more than an extension of polar coordinates into three dimensions. The following are the conversion formulas for cylindrical coordinates.

In order to do the integral in cylindrical coordinates we will need to know what \(dV\) will become in terms of cylindrical coordinates. We will be able to show in the Change of Variables section of this chapter that,

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