Finding Bounds for a Triple Integral in the First Octant

[harpazo]

Set up a triple integral for the volume of the solid. DO NOT FIND THE VOLUME.

The solid in the first octant bounded by the coordinates planes and the plane z = 4 - x - y.

We are dealing with the xy-plane where z = 0.

I know that 0 ≤ z ≤ 4 - x - y. The bounds of the integral pertaining to dz are z = 0 to z = 4 - x - y.

How do I find the other bounds?

 

[Prove It]

Set up a triple integral for the volume of the solid. DO NOT FIND THE VOLUME.

The solid in the first octant bounded by the coordinates planes and the plane z = 4 - x - y.

We are dealing with the xy-plane where z = 0.

I know that 0 ≤ z ≤ 4 - x - y. The bounds of the integral pertaining to dz are z = 0 to z = 4 - x - y.

How do I find the other bounds?

Since you're in the first octant you immediately know that all variables are nonnegative, so $\displaystyle \begin{align*} x, y, z \geq 0 \end{align*}$. When z = 0 we have $\displaystyle \begin{align*} y = 4 - x \end{align*}$. When x = 0, y is at its maximum of 4, and when y = 0, x is at its maximum of 4. Thus the bounds are $\displaystyle \begin{align*} 0 \leq z \leq 4 - x - y , \, 0 \leq y \leq 4 - x, \, 0 \leq x \leq 4 \end{align*}$.

 

[HallsofIvy]

The plane x+ y+ z= 4 cuts the xy-plane (z= 0) at x+ y= 4 or y= 4- x. That line cuts the x-axis (y= 0) at x= 4. x goes from 0 to 4, for each x, y goes from 0 to 4- x and, for each (x, y), z goes from 0 to 4- x- y. The integral is $$\int_0^4\int_0^{4-x}\int_0^{4-x-y} dzdydx$$.

 

[harpazo]

Since you're in the first octant you immediately know that all variables are nonnegative, so $\displaystyle \begin{align*} x, y, z \geq 0 \end{align*}$. When z = 0 we have $\displaystyle \begin{align*} y = 4 - x \end{align*}$. When x = 0, y is at its maximum of 4, and when y = 0, x is at its maximum of 4. Thus the bounds are $\displaystyle \begin{align*} 0 \leq z \leq 4 - x - y , \, 0 \leq y \leq 4 - x, \, 0 \leq x \leq 4 \end{align*}$.

Thanks for providing the bounds. I am not too clear on the first octant idea.

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I definitely need to practice finding the bounds more than I originally considered.

 

[MarkFL]

Thanks for providing the bounds. I am not too clear on the first octant idea...

Three perpedicular axes will divide the space into 8 regions known as octants:

The octant where all coordinates are positive is know as the first octant...in the image above, it is the top-front-right octant.

 

[harpazo]

Three perpedicular axes will divide the space into 8 regions known as octants:
The octant where all coordinates are positive is know as the first octant...in the image above, it is the top-front-right octant.

Nice picture. Good notes.

 

[harpazo]

I need to practice setting up double and triple integrals. Evaluating in most cases is not an issue.

 

[MarkFL]

I need to practice setting up double and triple integrals. Evaluating in most cases is not an issue.

Being able to change the order of integration can be crucial as well. Some iterated integrals that may be impossible to evaluate using one order of integration can, perhaps, be evaluated using the reverse order of integration.

Consider the following example:

Evaluate \(\displaystyle \iint\limits_{R}xe^{y^2}\,dA\) over the region bounded by $y=x^2$, $x=0$ and $y=4$.

When viewed as a Type I region (vertical strips), we find:

\(\displaystyle 0\le x\le2\)

\(\displaystyle x^2\le y\le4\)

And so:

\(\displaystyle \iint\limits_{R}xe^{y^2}\,dA=\int_0^2 x\int_{x^2}^4 e^{y^2}\,dy\,dx\)

The difficulty here is that the partial integral \(\displaystyle \int_{x^2}^4 e^{y^2}\,dy\) cannot be evaluated since $e^{y^2}$ has no elementary antiderivative with respect to $y$. However, we can interpret the same region as a Type II region (horizontal strips) defined by:

\(\displaystyle 0\le y\le4\)

\(\displaystyle 0\le x\le\sqrt{y}\)

Hence:

\(\displaystyle \iint\limits_{R}xe^{y^2}\,dA=\int_0^4 e^{y^2}\int_{0}^{\sqrt{y}}x \,dx\,dy=\frac{1}{4}\int_0^4 2ye^{y^2}\,dy=\frac{1}{4}\left(e^{16}-1\right)\)

 

[harpazo]

Being able to change the order of integration can be crucial as well. Some iterated integrals that may be impossible to evaluate using one order of integration can, perhaps, be evaluated using the reverse order of integration.

Consider the following example:

Evaluate \(\displaystyle \iint\limits_{R}xe^{y^2}\,dA\) over the region bounded by $y=x^2$, $x=0$ and $y=4$.

When viewed as a Type I region (vertical strips), we find:

\(\displaystyle 0\le x\le2\)

\(\displaystyle x^2\le y\le4\)

And so:

\(\displaystyle \iint\limits_{R}xe^{y^2}\,dA=\int_0^2 x\int_{x^2}^4 e^{y^2}\,dy\,dx\)

The difficulty here is that the partial integral \(\displaystyle \int_{x^2}^4 e^{y^2}\,dy\) cannot be evaluated since $e^{y^2}$ has no elementary antiderivative with respect to $y$. However, we can interpret the same region as a Type II region (horizontal strips) defined by:

\(\displaystyle 0\le y\le4\)

\(\displaystyle 0\le x\le\sqrt{y}\)

Hence:

\(\displaystyle \iint\limits_{R}xe^{y^2}\,dA=\int_0^4 e^{y^2}\int_{0}^{\sqrt{y}}x \,dx\,dy=\frac{1}{4}\int_0^4 2ye^{y^2}\,dy=\frac{1}{4}\left(e^{16}-1\right)\)

Please, explain the difference between vertical and horizontal strips via geometric interpretation.

 

[MarkFL]

Please, explain the difference between vertical and horizontal strips via geometric interpretation.

Consider the following diagram of the region $R$ from the example I posted:

View attachment 6626

Now, if we consider this a region of Type I, then the outer integral will be in terms of $x$, which we see ranges from $0\le x\le 2$, and the inner integral will be in terms of $y$, which we see ranges from $x^2\le y\le4$...the vertical red strip.

If we consider this a region of Type II, then the outer integral will be in terms of $y$, which we see ranges from $0\le y\le 4$, and the inner integral will be in terms of $x$, which we see ranges from $0\le x\le\sqrt{y}$...the horizontal green strip.

 

[harpazo]

Consider the following diagram of the region $R$ from the example I posted:
Now, if we consider this a region of Type I, then the outer integral will be in terms of $x$, which we see ranges from $0\le x\le 2$, and the inner integral will be in terms of $y$, which we see ranges from $x^2\le y\le4$...the vertical red strip.

If we consider this a region of Type II, then the outer integral will be in terms of $y$, which we see ranges from $0\le y\le 4$, and the inner integral will be in terms of $x$, which we see ranges from $0\le x\le\sqrt{y}$...the horizontal green strip.

An excellent reply.