How Do You Set Limits for a Triple Integral Over a Parabolic Cylinder?

A parabolic cylinder is a surface, not a complete bounded solid. Its equation describes a parabola in one coordinate plane that extends unchanged along a third direction. To set up a triple integral, you also need the other boundaries of the solid—such as planes, another cylinder, or coordinate planes.

Here is the standard approach, with a worked example for a cylinder extending parallel to the $z$-axis.

Write the cylinder as an inequality

A common parabolic cylinder is

$$y^2=4px,$$

where $p>0$. The equation is independent of $z$, so the surface extends in the $z$-direction. Its cross-section in the $xy$-plane is a parabola opening toward positive $x$.

If the solid is on the inside, or concave, side of the cylinder, the corresponding inequality is

$$x\geq \frac{y^2}{4p}.$$

That inequality gives only one boundary. You still need something to stop the region in the $x$-direction, such as $x=b$, and lower and upper surfaces in the $z$-direction, such as $z=c$ and $z=d$.

The axis and orientation may differ. For example, $x^2=4py$ describes a cylinder extending parallel to the $z$-axis but with a parabola opening toward positive $y$. A surface involving $z$ instead may extend along a different coordinate direction. Read the equation before assigning limits.

Choose an order from the region

Each integration order describes the same solid using a different sequence of bounds. A useful way to find limits is to project the solid onto a coordinate plane, then describe the variable being integrated first as lying between its lower and upper surfaces.

For a solid bounded by $y^2=4px$, $x=b$, and horizontal planes $z=c$ and $z=d$, with $b>0$ and $c<d$, the region is

$$\frac{y^2}{4p}\leq x\leq b,\qquad c\leq z\leq d.$$

The inequality $y^2/(4p)\leq b$ also gives the full range of $y$:

$$-2\sqrt{pb}\leq y\leq 2\sqrt{pb}.$$

Thus, for the order $dx\,dy\,dz$, a triple integral of a function $f$ is

$$\int_c^d\int_{-2\sqrt{pb}}^{2\sqrt{pb}}\int_{y^2/(4p)}^b f(x,y,z)\,dx\,dy\,dz.$$

Alternatively, solve the cylinder inequality for $y$. At each fixed $x$ between $0$ and $b$, $y$ ranges from $-2\sqrt{px}$ to $2\sqrt{px}$. The order $dy\,dx\,dz$ gives

$$\int_c^d\int_0^b\int_{-2\sqrt{px}}^{2\sqrt{px}} f(x,y,z)\,dy\,dx\,dz.$$

Both integrals cover the same solid. The better order is usually the one that makes the bounds and the integration simpler.

Worked example

Take the solid bounded by $y^2=4x$, $x=4$, $z=0$, and $z=3$, on the side $x\ge y^2/4$. Here $p=1$, $b=4$, $c=0$, and $d=3$.

For $dx\,dy\,dz$, the parabola and plane meet where $y^2=16$, so $-4\le y\le4$. At each fixed $y$, $x$ runs from the cylinder to the plane, and $z$ runs from 0 to 3:

$$\int_0^3\int_{-4}^{4}\int_{y^2/4}^{4} f(x,y,z)\,dx\,dy\,dz.$$

For $dy\,dx\,dz$, $x$ runs from 0 to 4; at each fixed $x$, $y$ runs between the two branches of the parabola:

$$\int_0^3\int_0^4\int_{-2\sqrt{x}}^{2\sqrt{x}} f(x,y,z)\,dy\,dx\,dz.$$

As a check, set $f=1$ to calculate the volume. Using the second order,

$$\begin{aligned} V&=\int_0^3\int_0^4\int_{-2\sqrt{x}}^{2\sqrt{x}}1\,dy\,dx\,dz\\ &=3\int_0^4 4\sqrt{x}\,dx\\ &=3\left[\frac{8}{3}x^{3/2}\right]_0^4=64. \end{aligned}$$

The result is consistent with the cross-sectional area: at height $z$, the $xy$-region has area $\int_0^4 4\sqrt{x}\,dx=64/3$, and its height is 3, giving volume 64.

Common mistakes

• Using the surface as if it were the whole solid. The equation $y^2=4px$ gives a boundary, not finite upper and lower limits in every direction. State the additional surfaces that enclose the region.
• Using only one branch of the parabola. When the solid includes both sides of the axis, the $y$-limits are negative and positive: $-2\sqrt{px}$ to $2\sqrt{px}$.
• Mixing up a surface equation and an inequality. The equation identifies the boundary; decide which side belongs to the solid from the problem’s description or a sketch.
• Assuming all orders have the same-looking bounds. Changing the order can make a bound variable-dependent or require splitting the region into pieces. Re-derive the bounds for each order rather than merely rearranging them.
• Leaving out the cylinder’s direction. A parabolic cylinder is constant along one direction. Determine which variable is absent from its equation before deciding how that coordinate is bounded.

If a problem gives only the words “parabolic cylinder” and no other boundary information, its triple-integral limits cannot be determined uniquely. Use the equation and all the bounding surfaces to specify the solid first.