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Triple Integrals and Solid Regions · Tutorial 614 of 999

Solids Between Curved Graphs

Determine where curved surfaces enclose a solid, express that solid with Cartesian bounds, and evaluate triple integrals over it.

Intermediate 9 min read

What You'll Learn

  • Identify the planar projection by comparing the lower and upper surface heights.
  • Convert intersections of curved surfaces into Cartesian bounds.
  • Set up an iterated triple integral for a solid between two graphs.
  • Evaluate integrals over rectangular and curved projections.
  • Detect invalid bounds when the proposed lower surface exceeds the upper one.

From Surface Equations to Cartesian Bounds

A solid between two surfaces is described by its vertical extent above each point in a planar region. The key task is to find that region correctly: a pair of surface equations alone does not tell us where the lower graph is actually below the upper graph.

Once the projection onto the \(xy\)-plane is known, the solid can be written using a lower bound for \(z\), an upper bound for \(z\), and Cartesian bounds for \(x\) and \(y\). The resulting iterated integral follows the same principles as other triple integrals, but the curved surfaces make identifying and checking the projection especially important.

Definition: A solid is vertically simple over a planar region \(D\) if it has the form \(E=\{(x,y,z):(x,y)\in D,\ L(x,y)\le z\le U(x,y)\}\), where \(L\) and \(U\) are its lower and upper surface functions and \(L(x,y)\le U(x,y)\) throughout \(D\).

For a vertically simple solid, an integral of a continuous function \(f\) can be written by integrating first with respect to \(z\), then over the projection \(D\):

$$ \iiint_E f(x,y,z)\,dV = \iint_D\left(\int_{L(x,y)}^{U(x,y)} f(x,y,z)\,dz\right)dA. $$

Here \(dA\) denotes planar area in the \(xy\)-plane. If \(D\) is described with \(x\)-limits and \(y\)-limits, then \(dA\) is represented by \(dy\,dx\) or \(dx\,dy\), according to those limits.

Finding the Projection

Suppose a solid lies between \(z=L(x,y)\) and \(z=U(x,y)\). At a fixed point \((x,y)\), the vertical segment belongs to the solid precisely when \(L(x,y)\le U(x,y)\). Therefore the projection is the set of points satisfying that inequality, along with any other planar restrictions imposed by the solid.

The equality \(L(x,y)=U(x,y)\) often gives the boundary of the projection. Solving that equality is not enough by itself: after finding a candidate region, check that the inequality \(L\le U\) holds throughout it.

1
Name the two graphs.
Write the proposed lower height \(L(x,y)\) and upper height \(U(x,y)\).
2
Compare their heights.
Solve \(L(x,y)\le U(x,y)\) to identify where vertical segments are possible.
3
Describe the planar region.
Use the inequality and any additional constraints to obtain valid Cartesian bounds for \(x\) and \(y\).
4
Write and evaluate the integral.
Integrate \(z\) from \(L(x,y)\) to \(U(x,y)\), then integrate over the planar projection.

Why the Height Difference Appears

For volume, the integrand is \(1\). At a fixed point \((x,y)\), integrating \(1\) along the vertical segment gives the segment's length. The Fundamental Theorem of Calculus makes this precise:

$$ \int_{L(x,y)}^{U(x,y)}1\,dz = \left[z\right]_{L(x,y)}^{U(x,y)} = U(x,y)-L(x,y). $$

Integrating this length over the projection adds the vertical contributions across the solid. Thus, when \(L\le U\) on \(D\), the volume is \(\iint_D(U-L)\,dA\). This also explains why the ordering of the surfaces matters: reversing their roles changes the sign of the height difference.

Derivation: For a continuous \(f\) on a bounded vertically simple solid, Fubini’s theorem permits integration along each vertical segment and then over the projection. Applying the Fundamental Theorem of Calculus to the inner integral gives \(\int_{L(x,y)}^{U(x,y)}f(x,y,z)\,dz=F(x,y,U(x,y))-F(x,y,L(x,y))\), where \(F\) is an antiderivative of \(f\) with respect to \(z\). For \(f=1\), take \(F=z\), yielding the height \(U-L\). The surface functions and planar bounds must describe the whole solid, and the integrand must meet the usual integrability conditions; continuity on a bounded solid is sufficient.

Worked Example: Two Paraboloid Graphs over a Square

Worked Example: Two Paraboloid Graphs over a Square

Find the volume of the solid over \(-1\le x\le1\), \(-1\le y\le1\), between the lower graph \(z=x^2+y^2\) and the upper graph \(z=x^2+y^2+2\).

The height difference is \((x^2+y^2+2)-(x^2+y^2)=2\), which is positive throughout the square. Thus the entire square is a valid projection. The bounds are \(-1\le x\le1\), \(-1\le y\le1\), and \(x^2+y^2\le z\le x^2+y^2+2\). Integrating in the order \(dz\,dy\,dx\) gives

$$ \begin{aligned} V &=\int_{-1}^{1}\int_{-1}^{1}\int_{x^2+y^2}^{x^2+y^2+2}1\,dz\,dy\,dx\\ &=\int_{-1}^{1}\int_{-1}^{1} \left[(z)\right]_{x^2+y^2}^{x^2+y^2+2}\,dy\,dx\\ &=\int_{-1}^{1}\int_{-1}^{1}2\,dy\,dx\\ &=\int_{-1}^{1}4\,dx\\ &=8. \end{aligned} $$

Check: The vertical height is \(2\) everywhere, and the square has area \(2\cdot2=4\). Height times area is \(2\cdot4=8\), agreeing with the integral.

Worked Example: A Paraboloid Below a Plane

Worked Example: A Paraboloid Below a Plane

Find the volume bounded below by \(z=x^2+y^2\) and above by \(z=2x\).

The surfaces meet where \(x^2+y^2=2x\). Completing the square gives \((x-1)^2+y^2=1\). The inequality for the lower graph to lie below the upper graph is \(x^2+y^2\le2x\), or \((x-1)^2+y^2\le1\). This disk is the projection. Describing it with \(y\)-bounds gives \(0\le x\le2\) and \(-\sqrt{2x-x^2}\le y\le\sqrt{2x-x^2}\). The volume integral is

$$ V=\int_0^2\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} \int_{x^2+y^2}^{2x}1\,dz\,dy\,dx. $$

Let \(A=2x-x^2\) for the following simplification. The inner integral is \(2x-x^2-y^2=A-y^2\), so

$$ \begin{aligned} V &=\int_0^2\int_{-\sqrt A}^{\sqrt A}(A-y^2)\,dy\,dx\\ &=\int_0^2\left[Ay-\frac{y^3}{3}\right]_{-\sqrt A}^{\sqrt A}\,dx\\ &=\int_0^2\left(2A^{3/2}-\frac{2}{3}A^{3/2}\right)\,dx\\ &=\frac43\int_0^2(2x-x^2)^{3/2}\,dx. \end{aligned} $$

To evaluate the last integral, set \(u=x-1\). Then \(2x-x^2=1-u^2\), and the limits become \(-1\) and \(1\). Next set \(u=\sin t\), so \(du=\cos t\,dt\) and \((1-u^2)^{3/2}=\cos^3t\) on \(-\pi/2\le t\le\pi/2\). Therefore

$$ \begin{aligned} \int_0^2(2x-x^2)^{3/2}\,dx &=\int_{-1}^{1}(1-u^2)^{3/2}\,du\\ &=\int_{-\pi/2}^{\pi/2}\cos^4t\,dt\\ &=\int_{-\pi/2}^{\pi/2}\frac{3+4\cos(2t)+\cos(4t)}{8}\,dt\\ &=\frac{3\pi}{8}. \end{aligned} $$

The integrals of \(\cos(2t)\) and \(\cos(4t)\) over these limits are zero, since their antiderivatives have equal endpoint values. Hence \(V=\frac43\cdot\frac{3\pi}{8}=\frac{\pi}{2}\).

Check: On the projection, the height \(2x-x^2-y^2\) is nonnegative by the disk inequality, so the integral represents a nonnegative volume. Also, differentiating \(Ay-y^3/3\) with respect to \(y\) gives \(A-y^2\), the height integrand used in the calculation.

Worked Example: Integrating a Nonconstant Function

Worked Example: Integrating a Nonconstant Function

Evaluate \(\iiint_E z\,dV\), where \(E\) lies over the square \(-1\le x\le1\), \(-1\le y\le1\), between \(z=x^2\) and \(z=2-y^2\).

The proposed height is \(2-y^2-x^2=2-x^2-y^2\). On the square, \(x^2\le1\) and \(y^2\le1\), so \(2-x^2-y^2\ge0\). The surfaces are correctly ordered everywhere, with equality at the corners. Set up the integral in the order \(dz\,dx\,dy\):

$$ \begin{aligned} \iiint_E z\,dV &=\int_{-1}^{1}\int_{-1}^{1}\int_{x^2}^{2-y^2}z\,dz\,dx\,dy\\ &=\int_{-1}^{1}\int_{-1}^{1} \frac12\left((2-y^2)^2-x^4\right)\,dx\,dy. \end{aligned} $$

Integrate first with respect to \(x\). Since \(\int_{-1}^1 1\,dx=2\) and \(\int_{-1}^1x^4\,dx=2/5\),

$$ \begin{aligned} \iiint_E z\,dV &=\int_{-1}^{1} \left((2-y^2)^2-\frac15\right)\,dy\\ &=\int_{-1}^{1} \left(\frac{19}{5}-4y^2+y^4\right)\,dy\\ &=\left[\frac{19}{5}y-\frac43y^3+\frac15y^5\right]_{-1}^{1}\\ &=\frac{38}{5}-\frac83+\frac25\\ &=8-\frac83\\ &=\frac{16}{3}. \end{aligned} $$

Check: The antiderivative in the last line differentiates to \(\frac{19}{5}-4y^2+y^4\), which equals \((2-y^2)^2-\frac15\). The integrand \(z\) is nonnegative throughout \(E\), since its lower bound is \(x^2\ge0\); the positive result is consistent with that fact.

A Common Error: Using the Wrong Projection

For the paraboloid-and-plane example, the correct projection is \((x-1)^2+y^2\le1\). It may be tempting to replace it with the convenient rectangle \(0\le x\le2\), \(-1\le y\le1\), but that rectangle contains points where the proposed lower surface is higher than the proposed upper surface.

For instance, the point \((x,y)=(\frac12,1)\) lies inside that rectangle. At this point the proposed lower height is \(x^2+y^2=\frac14+1=\frac54\), while the proposed upper height is \(2x=1\). Thus \(\frac54>1\): there is no vertical segment between those surfaces in the claimed order. In fact, this point is outside the correct projection because \((\frac12-1)^2+1^2=\frac54>1\).

Correction: Do not choose planar limits merely because they are simple. Derive the projection from \(L(x,y)\le U(x,y)\), then verify that the resulting Cartesian bounds cover exactly that region.

Why These Bounds Matter

Curved graphs can meet along a curve whose projection has curved edges, as in the disk in the paraboloid-and-plane example. Other surfaces may enclose a solid over a rectangle or another planar region. In each case, the geometry determines the limits: the \(z\)-bounds specify the vertical segment, while the \(x\)- and \(y\)-bounds specify where that segment exists.

Before evaluating an integral, check three things:

  • The chosen lower surface is no higher than the chosen upper surface throughout the stated projection.
  • The Cartesian limits describe the entire projection and do not include extra points where the surface order fails.
  • The integrand is integrated using the same bounds and order that define the solid.

These checks are particularly useful when an intersection produces a curved planar boundary. A correct iterated integral is a mathematical description of the solid, not merely a sequence of integration signs.

Check Your Understanding

Use the surface comparison and the Cartesian bounds to answer these questions.

  1. For \(z=x^2+y^2\) below \(z=3\), what inequality describes the projection onto the \(xy\)-plane?
  2. For \(z=x^2+y^2\) below \(z=2x\), complete the square in the intersection equation and state the projected disk.
  3. Why is the rectangle \(0\le x\le2\), \(-1\le y\le1\) invalid as the full projection in the paraboloid-and-plane example?
  4. Write an iterated integral for the volume between \(z=x^2\) and \(z=2-y^2\) over \(-1\le x\le1\), \(-1\le y\le1\).
  5. For a solid between \(z=L(x,y)\) and \(z=U(x,y)\), what does the inner integral of \(1\) with respect to \(z\) represent?

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