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17calculus > partial integrals > double integrals area & volume

In single variable calculus, you learned how to calculate area between curves and volumes of revolution. Now that you know how to evaluate double integrals, you can use that knowledge to calculate the same areas and also volumes of more irregular shapes that lie above a plane, usually the xy-plane.
This page covers area and volume in rectangular coordinates. Double integrals in polar coordinates are covered on a separate page.

Actually, you already know how to calculate area and volume with a double integral that is already set up. You can think of the limits of integration as defining the area in the plane. To find that area, the integrand is just one. For volume, the integrand is the height of the volume above that area. So on this page, we show you how to set up the double integral to calculate an area and a volume.

In your classes and in many of the videos you will watch, the instructor will often generate a 3D plot of the volume you are trying to calculate. These are interesting and can give you some perspective on what is going on but we have found them unnecessary when actually setting up the integral. The only plotting you really need to do is 2D in the base plane, i.e. the plane where the shadow of the volume falls. This is easily done with winplot or other graphing utilities and can usually also be done by hand. This 2D plot is a necessary part of setting up your double integral since you use it not only to determine your limits of integration but also in what order you integrate.

Whether you are calculating an area or a volume, you start out by describing the area in the plane. Just like you did when calculating volumes of revolution using either the washer or cylinder method, you need to plot the region in the base plane and describe it mathematically. This panel explains how to do that.

Describing A Region In The xy-Plane

To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot ( used to plot these graphs; we used gimp to add labels and other graphics ). However, graphing by hand is usually the best and quickest way.

We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by $$f(x)$$ (red line), $$g(x)$$ (blue line) and $$x=a$$ (black line).
[Remember that an equation like $$x=a$$ can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]

Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).

Vertically

Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.

Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing $$g(x)$$, no matter where we draw it. Similarly, the arrow always exits the area by crossing $$f(x)$$, no matter where we draw it. Do you see that?

But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions $$f(x)$$ and $$g(x)$$ intersect. You should be able to do that. We will call that point $$(b,f(b))$$. Also, we will call the left boundary $$x=a$$. So now we have everything we need to describe this area. We give the final results below.

Vertical Arrow

$$g(x) \leq y \leq f(x)$$

arrow leaves through $$f(x)$$ and enters through $$g(x)$$

$$a \leq x \leq b$$

arrow sweeps from left ($$x=a$$) to right ($$x=b$$)

Horizontally

We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of $$f(x)$$ and $$g(x)$$ in terms of $$y$$. ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations $$f(x) \to F(y)$$ and $$g(x) \to G(y)$$.

Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line $$x=a$$. However, there is something strange going on at the point $$(b,f(b))$$. Notice that when the arrow is below $$f(b)$$, the arrow exits through $$g(x)$$ but when the arrow is above $$f(b)$$, the arrow exits through $$f(x)$$. This is a problem. To overcome this, we need to break the area into two parts at $$f(b)$$.

Lower Section - - This section is described by the arrow leaving through $$g(x)$$. So the arrow sweeps from $$g(a)$$ to $$g(b)$$.
Upper Section - - This section is described by the arrow leaving through $$f(x)$$. The arrow sweeps from $$f(b)$$ to $$f(a)$$.
The total area is the combination of these two areas. The results are summarized below.

Horizontal Arrow

lower section

$$a \leq x \leq G(y)$$

arrow leaves through $$G(y)$$ and enters through $$x=a$$

$$g(a) \leq y \leq g(b)$$

arrow sweeps from bottom ($$y=g(a)$$) to top ($$y=g(b)$$)

upper section

$$a \leq x \leq F(y)$$

arrow leaves through $$F(y)$$ and enters through $$x=a$$

$$f(b) \leq y \leq f(a)$$

arrow sweeps from bottom ($$y=f(b)$$) to top ($$y=f(a)$$)

Type 1 and Type 2 Regions

Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.

Here is a quick video clip going into more detail on Type 1 and Type 2 regions.

Krista King Math - type I and type 2 regions [1min-39secs]

video by Krista King Math

[ For some videos and practice problems dedicated to this topic, check out this page. ]

Okay, so now that we have a description of the area in the plane, we need to set up the integral to calculate either the area or the volume. From the above panel, we know the limits of integration. The order of integration is as follows.
Outside Integral - - The outside integral is the one with the constant limits of integration and the sweeping of the arrow.
Inside Integral - - The inside integral is the other set of limits and the part of the arrow that is in the shaded region.
Here are the integrals.

 if the arrow is vertical $$\displaystyle{\int_{a}^{b}{ \int_{g(x)}^{f(x)}{ h(x,y) ~dy } ~dx }}$$ if the arrow is horizontal $$\displaystyle{\int_{c}^{d}{ \int_{g(y)}^{f(y)}{ h(x,y) ~dx } ~dy }}$$

Okay, so now the integral is almost complete. The only thing left is to determine the integrand, $$h(x,y)$$.

$$h(x,y)$$

integral

to calculate area in the plane

$$h(x,y) = 1$$

$$\displaystyle{ \iint\limits_R {1 ~ dA} }$$

to calculate surface area

$$h(x,y) = \sqrt{ 1+(\partial z/ \partial x)^2 + (\partial z/ \partial y)^2 }$$

$$\displaystyle{ \iint\limits_R {\sqrt{ 1+(\partial z/ \partial x)^2 + (\partial z/ \partial y)^2 } ~ dA} }$$

to calculate volume

$$h(x,y)$$ is the height above the xy-plane

$$\displaystyle{ \iint\limits_R {h(x,y) ~ dA} }$$

This video clip explains calculation of volumes with double integrals in more detail.

Dr Chris Tisdell - calculate volume with double integrals [9mins]

video by Dr Chris Tisdell

Practice - Calculate Area in the Plane

Okay, let's work some practice problems. Let's start with calculating areas in the plane.

Calculate the area of the region bounded by $$y = x$$ and $$y = x^2$$

Problem Statement

Calculate the area of the region bounded by $$y = x$$ and $$y = x^2$$

Solution

739 solution video

video by Dr Chris Tisdell

Calculate the area of the region bounded by the curves $$y = \sqrt{x}, y = x, y = x/2$$

Problem Statement

Calculate the area of the region bounded by the curves $$y = \sqrt{x}, y = x, y = x/2$$

Solution

742 solution video

video by Dr Chris Tisdell

Practice - Calculate Surface Area

Okay, if you are comfortable working those double integrals, let's try finding surface area. To remind you, the integral you need is $$\displaystyle{ \iint\limits_R {\sqrt{ 1+(\partial z/ \partial x)^2 + (\partial z/ \partial y)^2 } ~ dA} }$$.

Find the area of the surface of the paraboloid $$z = x^2 + y^2$$ that lies below the plane $$z = 16$$.

Problem Statement

Find the area of the surface of the paraboloid $$z = x^2 + y^2$$ that lies below the plane $$z = 16$$.

Solution

2301 solution video

video by MIP4U

Find the surface area of the part of the plane $$z = 3 + 2x + 4y$$ that lies inside the cylinder $$x^2 + y^2 = 4$$.

Problem Statement

Find the surface area of the part of the plane $$z = 3 + 2x + 4y$$ that lies inside the cylinder $$x^2 + y^2 = 4$$.

Solution

2308 solution video

video by MIP4U

Practice - Calculate Volume

Time for some volume practice problems. To remind you, the integral you need is $$\displaystyle{ \iint\limits_R {h(x,y) ~ dA} }$$ where $$h(x,y)$$ is the height above the xy-plane.

Instructions - - Unless otherwise instructed, use a double integral to calculate the volume above the specified area, giving your answers in exact terms.

Basic Problems

$$f(x,y) = x+y; R = { (x,y): ~0 \leq x \leq 1, ~x^2 \leq y \leq x }$$

Problem Statement

$$f(x,y) = x+y; R = { (x,y): ~0 \leq x \leq 1, ~x^2 \leq y \leq x }$$

Solution

736 solution video

video by Dr Chris Tisdell

$$z = 4-x-y; R = \{(x,y) : ~0 \leq x \leq 2, ~0 \leq y \leq 1\}$$

Problem Statement

$$z = 4-x-y; R = \{(x,y) : ~0 \leq x \leq 2, ~0 \leq y \leq 1\}$$

Solution

740 solution video

video by Dr Chris Tisdell

$$f(x,y) = 1 - 6x^2y; R = \{ (x,y) : 0 \leq x \leq 2, -1 \leq y \leq 1 \}$$

Problem Statement

$$f(x,y) = 1 - 6x^2y; R = \{ (x,y) : 0 \leq x \leq 2, -1 \leq y \leq 1 \}$$

Solution

741 solution video

video by Dr Chris Tisdell

$$x^2+(1/2)y^2 + z = 12; 0 \leq x \leq 2, 0 \leq y \leq 3$$

Problem Statement

$$x^2+(1/2)y^2 + z = 12; 0 \leq x \leq 2, 0 \leq y \leq 3$$

Solution

744 solution video

video by MIP4U

$$f(x,y) = xy^2; R:[1,2] \times [0,3]$$

Problem Statement

$$f(x,y) = xy^2; R:[1,2] \times [0,3]$$

Solution

745 solution video

video by MIP4U

$$f(x,y) = x\cos(xy); R:[0,1] \times [0, \pi/2]$$

Problem Statement

$$f(x,y) = x\cos(xy); R:[0,1] \times [0, \pi/2]$$

Solution

746 solution video

video by MIP4U

$$z = x + y ; 0 \leq x \leq y/2, ~ 0 \leq y \leq 6$$

Problem Statement

$$z = x + y ; 0 \leq x \leq y/2, ~ 0 \leq y \leq 6$$

Solution

747 solution video

video by MIP4U

$$f(x,y) = xy ; 0 \leq x \leq 1, ~ x^2 \leq y \leq x$$

Problem Statement

$$f(x,y) = xy ; 0 \leq x \leq 1, ~ x^2 \leq y \leq x$$

Solution

748 solution video

video by Dr Chris Tisdell

$$z = xy$$; triangle with vertices $$(1,1), (4,1), (1,2)$$

Problem Statement

$$z = xy$$; triangle with vertices $$(1,1), (4,1), (1,2)$$

Solution

1190 solution video

video by Krista King Math

$$z = y\sqrt{x^3+1} ; \{(x,y)~:~ 0 \leq y \leq x \leq 4 \}$$

Problem Statement

$$z = y\sqrt{x^3+1} ; \{(x,y)~:~ 0 \leq y \leq x \leq 4 \}$$

Solution

1906 solution video

video by MIP4U

Intermediate Problems

$$h(x,y) = 3-x-y$$; region bounded by $$y=0, ~x=1, ~y=x$$

Problem Statement

$$h(x,y) = 3-x-y$$; region bounded by $$y=0, ~x=1, ~y=x$$

Solution

737 solution video

video by Dr Chris Tisdell

$$\sin(x^2)$$; region bounded by $$x = y, ~x = 2$$ and the y-axis

Problem Statement

$$\sin(x^2)$$; region bounded by $$x = y, ~x = 2$$ and the y-axis

Solution

738 solution video

video by Dr Chris Tisdell

Calculate the volume bounded by the coordinate planes and $$z = 2-2x-y$$

Problem Statement

Calculate the volume bounded by the coordinate planes and $$z = 2-2x-y$$

Solution

743 solution video

video by Dr Chris Tisdell

$$f(x,y) = y^2$$; region bounded by $$x \geq 0, ~ y = x-1, ~ y = x/2$$

Problem Statement

$$f(x,y) = y^2$$; region bounded by $$x \geq 0, ~ y = x-1, ~ y = x/2$$

Solution

751 solution video

video by MIP4U

$$z = 3xy - x^3$$; region bounded by $$y = x^3$$ and $$y = \sqrt{x}$$

Problem Statement

$$z = 3xy - x^3$$; region bounded by $$y = x^3$$ and $$y = \sqrt{x}$$

Solution

752 solution video

video by Dr Chris Tisdell

Calculate the volume enclosed by $$x^2+y^2 = 16 ; (x^2+y^2)/8 \leq z \leq 5$$.

Problem Statement

Calculate the volume enclosed by $$x^2+y^2 = 16 ; (x^2+y^2)/8 \leq z \leq 5$$.

Solution

773 solution video

video by Dr Chris Tisdell

A cylindrical drill with a radius of 5cm is used to bore a hole through the center of a sphere with a radius of 7cm. Find the volume of the ring-shaped solid that remains.

Problem Statement

A cylindrical drill with a radius of 5cm is used to bore a hole through the center of a sphere with a radius of 7cm. Find the volume of the ring-shaped solid that remains.

$$64 \pi \sqrt{6}$$ cm3

Problem Statement

A cylindrical drill with a radius of 5cm is used to bore a hole through the center of a sphere with a radius of 7cm. Find the volume of the ring-shaped solid that remains.

Solution

1922 solution video

video by MIP4U

$$64 \pi \sqrt{6}$$ cm3