A cylinder is a geometric body bounded by two parallel planes and a cylindrical surface. In the article, we will talk about how to find the area of a cylinder and, using the formula, we will solve several problems for example.
A cylinder has three surfaces: a top, a bottom, and a side surface.
The top and bottom of the cylinder are circles and are easy to define.
It is known that the area of a circle is equal to πr 2 . Therefore, the formula for the area of two circles (top and bottom of the cylinder) will look like πr 2 + πr 2 = 2πr 2 .
The third, side surface of the cylinder, is the curved wall of the cylinder. In order to better represent this surface, let's try to transform it to get a recognizable shape. Imagine that a cylinder is an ordinary tin can that does not have a top lid and bottom. Let's make a vertical incision on the side wall from the top to the bottom of the jar (Step 1 in the figure) and try to open (straighten) the resulting figure as much as possible (Step 2).
After the full disclosure of the resulting jar, we will see a familiar figure (Step 3), this is a rectangle. The area of a rectangle is easy to calculate. But before that, let us return for a moment to the original cylinder. The vertex of the original cylinder is a circle, and we know that the circumference of a circle is calculated by the formula: L = 2πr. It is marked in red in the figure.
When the side wall of the cylinder is fully expanded, we see that the circumference becomes the length of the resulting rectangle. The sides of this rectangle will be the circumference (L = 2πr) and the height of the cylinder (h). The area of a rectangle is equal to the product of its sides - S = length x width = L x h = 2πr x h = 2πrh. As a result, we have obtained a formula for calculating the lateral surface area of a cylinder.
The formula for the area of the lateral surface of a cylinder
S side = 2prh
Full surface area of a cylinder
Finally, if we add up the area of all three surfaces, we get the formula for the total surface area of a cylinder. The surface area of the cylinder is equal to the area of the top of the cylinder + the area of the base of the cylinder + the area of the side surface of the cylinder or S = πr 2 + πr 2 + 2πrh = 2πr 2 + 2πrh. Sometimes this expression is written by the identical formula 2πr (r + h).
The formula for the total surface area of a cylinder
S = 2πr 2 + 2πrh = 2πr(r + h)
r is the radius of the cylinder, h is the height of the cylinder
Examples of calculating the surface area of a cylinder
To understand the above formulas, let's try to calculate the surface area of a cylinder using examples.
1. The radius of the base of the cylinder is 2, the height is 3. Determine the area of the side surface of the cylinder.
The total surface area is calculated by the formula: S side. = 2prh
S side = 2 * 3.14 * 2 * 3
S side = 6.28 * 6
S side = 37.68
The lateral surface area of the cylinder is 37.68.
2. How to find the surface area of a cylinder if the height is 4 and the radius is 6?
The total surface area is calculated by the formula: S = 2πr 2 + 2πrh
S = 2 * 3.14 * 6 2 + 2 * 3.14 * 6 * 4
S = 2 * 3.14 * 36 + 2 * 3.14 * 24
Exist a large number of cylinder related tasks. In them, you need to find the radius and height of the body or the type of its section. Plus, sometimes you need to calculate the area of a cylinder and its volume.
What body is a cylinder?
In the course of the school curriculum, a circular, that is, a cylinder that is such at the base, is studied. But they also distinguish the elliptical appearance of this figure. From the name it is clear that its base will be an ellipse or oval.
The cylinder has two bases. They are equal to each other and are connected by segments that combine the corresponding points of the bases. They are called cylinder generators. All generators are parallel to each other and equal. They make up the lateral surface of the body.
In general, a cylinder is an inclined body. If the generators make a right angle with the bases, then they already speak of a straight figure.
Interestingly, a circular cylinder is a body of revolution. It is obtained by rotating a rectangle around one of its sides.
The main elements of the cylinder
The main elements of the cylinder are as follows.
- Height. It is the shortest distance between the bases of the cylinder. If it is straight, then the height coincides with the generatrix.
- Radius. Coincides with the one that can be carried out in the base.
- Axis. This is a straight line that contains the centers of both bases. The axis is always parallel to all generators. In a right cylinder, it is perpendicular to the bases.
- Axial section. It is formed when the cylinder intersects the plane containing the axis.
- Tangent plane. It passes through one of the generators and is perpendicular to the axial section, which is drawn through this generatrix.
How is a cylinder related to a prism inscribed in it or circumscribed near it?
Sometimes there are problems in which it is necessary to calculate the area of a cylinder, while some elements of the prism associated with it are known. How are these figures related?
If a prism is inscribed in a cylinder, then its bases are equal polygons. Moreover, they are inscribed in the corresponding bases of the cylinder. The side edges of the prism coincide with the generators.
The described prism has regular polygons at its bases. They are described near the circles of the cylinder, which are its bases. The planes that contain the faces of the prism touch the cylinder along the generators.
On the area of the lateral surface and base for a right circular cylinder
If you unfold the side surface, you get a rectangle. Its sides will coincide with the generatrix and the circumference of the base. Therefore, the lateral area of \u200b\u200bthe cylinder will be equal to the product of these two quantities. If you write the formula, you get the following:
S side \u003d l * n,
where n is the generatrix, l is the circumference.
Moreover, the last parameter is calculated by the formula:
l = 2 π*r,
here r is the radius of the circle, π is the number "pi", equal to 3.14.
Since the base is a circle, its area is calculated using the following expression:
S main \u003d π * r 2.
On the area of the entire surface of a right circular cylinder
Since it is formed by two bases and a lateral surface, these three quantities must be added. That is, the total area of the cylinder will be calculated by the formula:
S floor = 2 π * r * n + 2 π * r 2 .
It is often written in a different form:
S floor = 2 π * r (n + r).
On the areas of an inclined circular cylinder
As for the bases, all the formulas are the same, because they are still circles. But the side surface no longer gives a rectangle.
To calculate the side surface area of an inclined cylinder, you will need to multiply the values of the generatrix and the perimeter of the section, which will be perpendicular to the selected generatrix.
The formula looks like this:
S side \u003d x * P,
where x is the length of the generatrix of the cylinder, P is the perimeter of the section.
The cross section, by the way, is better to choose such that it forms an ellipse. Then the calculations of its perimeter will be simplified. The length of the ellipse is calculated using a formula that gives an approximate answer. But it is often enough for the tasks of the school course:
l \u003d π * (a + b),
where "a" and "b" are the semiaxes of the ellipse, that is, the distances from the center to its nearest and farthest points.
The area of the entire surface must be calculated using the following expression:
S floor = 2 π * r 2 + x * R.
What are some sections of a right circular cylinder?
When the section passes through the axis, then its area is determined as the product of the generatrix and the diameter of the base. This is due to the fact that it has the form of a rectangle, the sides of which coincide with the designated elements.
To find the cross-sectional area of a cylinder that is parallel to the axial one, you will also need a formula for a rectangle. In this situation, one of its sides will still coincide with the height, and the other will be equal to the chord of the base. The latter coincides with the section line along the base.
When the section is perpendicular to the axis, then it looks like a circle. Moreover, its area is the same as at the base of the figure.
It is also possible to intersect at some angle to the axis. Then in the section an oval or part of it is obtained.
Task examples
Task number 1. A straight cylinder is given, the base area of which is 12.56 cm 2 . It is necessary to calculate the total area of the cylinder if its height is 3 cm.
Decision. It is necessary to use the formula for the total area of a circular right cylinder. But it lacks data, namely the radius of the base. But the area of the circle is known. From it it is easy to calculate the radius.
It turns out to be equal to the square root of the quotient, which is obtained by dividing the base area by pi. Dividing 12.56 by 3.14 is 4. The square root of 4 is 2. Therefore, the radius will have exactly this value.
Answer: S floor \u003d 50.24 cm 2.
Task number 2. A cylinder with a radius of 5 cm is cut off by a plane parallel to the axis. The distance from the section to the axis is 3 cm. The height of the cylinder is 4 cm. It is required to find the area of the section.
Decision. The section shape is rectangular. One of its sides coincides with the height of the cylinder, and the other is equal to the chord. If the first value is known, then the second must be found.
To do this, you need to make an additional construction. At the base we draw two segments. Both of them will start at the center of the circle. The first will end in the center of the chord and equal to the known distance to the axis. The second is at the end of the chord.
You get a right triangle. The hypotenuse and one of the legs are known in it. The hypotenuse is the same as the radius. The second leg is equal to half the chord. The unknown leg, multiplied by 2, will give the required chord length. Let's calculate its value.
In order to find the unknown leg, you need to square the hypotenuse and the known leg, subtract the second from the first and take the square root. The squares are 25 and 9. Their difference is 16. After extracting the square root, 4 remains. This is the desired leg.
The chord will be equal to 4 * 2 = 8 (cm). Now you can calculate the cross-sectional area: 8 * 4 \u003d 32 (cm 2).
Answer: S sec is 32 cm 2.
Task number 3. It is necessary to calculate the area of the axial section of the cylinder. It is known that a cube with an edge of 10 cm is inscribed in it.
Decision. The axial section of the cylinder coincides with a rectangle that passes through the four vertices of the cube and contains the diagonals of its bases. The side of the cube is the generatrix of the cylinder, and the diagonal of the base coincides with the diameter. The product of these two quantities will give the area that you need to find out in the problem.
To find the diameter, you will need to use the knowledge that the base of the cube is a square, and its diagonal forms an equilateral right triangle. Its hypotenuse is the required diagonal of the figure.
To calculate it, you need the formula of the Pythagorean theorem. You need to square the side of the cube, multiply it by 2 and take the square root. Ten to the second power is one hundred. Multiplied by 2 is two hundred. The square root of 200 is 10√2.
The section is again a rectangle with sides 10 and 10√2. Its area is easy to calculate by multiplying these values.
Answer. S sec \u003d 100√2 cm 2.
How to calculate the surface area of a cylinder is the topic of this article. In any mathematical problem, you need to start with data entry, determine what is known and what to operate on in the future, and only then proceed directly to the calculation.
This three-dimensional body is a geometric figure of a cylindrical shape, bounded above and below by two parallel planes. If you apply a little imagination, you will notice that a geometric body is formed by rotating a rectangle around an axis, with the axis being one of its sides.
It follows from this that the described curve above and below the cylinder will be a circle, the main indicator of which is the radius or diameter.
Cylinder Surface Area - Online Calculator
This function finally facilitates the calculation process, and everything comes down to automatic substitution of the given values of the height and radius (diameter) of the base of the figure. The only thing that is required is to accurately determine the data and not make mistakes when entering numbers.
Cylinder side surface area
First you need to imagine how the sweep looks in two-dimensional space.
This is nothing more than a rectangle, one side of which is equal to the circumference. Its formula has been known since time immemorial - 2π *r, where r is the radius of the circle. The other side of the rectangle is equal to the height h. It won't be hard to find what you're looking for.
Sside= 2π *r*h,
where number π = 3.14.
Full surface area of a cylinder
To find the total area of the cylinder, you need to get S side add the areas of two circles, the top and bottom of the cylinder, which are calculated by the formula S o =2π*r2.
The final formula looks like this:
Sfloor\u003d 2π * r 2+ 2π*r*h.
Cylinder area - formula in terms of diameter
To facilitate calculations, it is sometimes necessary to make calculations through the diameter. For example, there is a piece of a hollow pipe of known diameter.
Without bothering with unnecessary calculations, we have a ready-made formula. Algebra for 5th grade comes to the rescue.
Sgender = 2π*r 2 + 2 π*r*h= 2 π*d 2 /4 + 2 π*h*d/2 = π *d 2 /2 + π *d*h,
Instead of r in the full formula you need to insert the value r=d/2.
Examples of calculating the area of a cylinder
Armed with knowledge, let's get down to practice.
Example 1 It is necessary to calculate the area of a truncated piece of pipe, that is, a cylinder.
We have r = 24 mm, h = 100 mm. You need to use the formula in terms of the radius:
S floor \u003d 2 * 3.14 * 24 2 + 2 * 3.14 * 24 * 100 \u003d 3617.28 + 15072 \u003d 18689.28 (mm 2).
We translate into the usual m 2 and get 0.01868928, approximately 0.02 m 2.
Example 2 It is required to find out the area of the inner surface of the asbestos stove pipe, the walls of which are lined with refractory bricks.
The data are as follows: diameter 0.2 m; height 2 m. We use the formula through the diameter:
S floor \u003d 3.14 * 0.2 2 / 2 + 3.14 * 0.2 * 2 \u003d 0.0628 + 1.256 \u003d 1.3188 m 2.
Example 3 How to find out how much material is needed to sew a bag, r \u003d 1 m and a height of 1 m.
One moment, there is a formula:
S side \u003d 2 * 3.14 * 1 * 1 \u003d 6.28 m 2.
Conclusion
At the end of the article, the question arose: are all these calculations and translations of one value into another really necessary? Why is all this necessary and most importantly, for whom? But do not neglect and forget simple formulas from high school.
The world has stood and will stand on elementary knowledge, including mathematics. And, when embarking on some important work, it is never superfluous to refresh the data of calculations in memory, applying them in practice with great effect. Accuracy - the politeness of kings.
Find the area of the axial section perpendicular to the bases of the cylinder. One of the sides of this rectangle is equal to the height of the cylinder, the other is equal to the diameter of the base circle. Accordingly, the cross-sectional area in this case will be equal to the product of the sides of the rectangle. S=2R*h, where S is the cross-sectional area, R is the radius of the base circle, given by the conditions of the problem, and h is the height of the cylinder, also given by the conditions of the problem.
If the section is perpendicular to the bases, but does not pass through the axis of rotation, the rectangle will not equal the diameter of the circle. It needs to be calculated. To do this, the task must say at what distance from the axis of rotation the section plane passes. For the convenience of calculations, construct a circle of the base of the cylinder, draw a radius and set aside on it the distance at which the section is located from the center of the circle. From this point, draw to the perpendiculars until they intersect with the circle. Connect the intersection points to the center. You need to find chords. Find the size of half a chord using the Pythagorean theorem. It will be equal to the square root of the difference of the squares of the radius of the circle from the center to the section line. a2=R2-b2. The whole chord will be, respectively, equal to 2a. Calculate the cross-sectional area, which is equal to the product of the sides of the rectangle, that is, S=2a*h.
The cylinder can be dissected without passing through the plane of the base. If the cross section is perpendicular to the axis of rotation, then it will be a circle. Its area in this case is equal to the area of the bases, that is, it is calculated by the formula S \u003d πR2.
Helpful advice
To more accurately imagine the section, make a drawing and additional constructions to it.
Sources:
- cylinder cross section area
The line of intersection of a surface with a plane belongs both to the surface and to the secant plane. The line of intersection of a cylindrical surface with a secant plane parallel to the straight generatrix is a straight line. If the cutting plane is perpendicular to the axis of the surface of revolution, the section will have a circle. In general, the line of intersection of a cylindrical surface with a cutting plane is a curved line.
You will need
- Pencil, ruler, triangle, patterns, compasses, measuring instrument.
Instruction
On the frontal projection plane P₂, the section line coincides with the projection of the secant plane Σ₂ in the form of a straight line.
Designate the points of intersection of the generatrices of the cylinder with the projection Σ₂ 1₂, 2₂, etc. to points 10₂ and 11₂.
On the plane P₁ is a circle. Points 1₂ , 2₂ marked on the section plane Σ₂, etc. with the help of a projection line, the connections will be projected on the outline of this circle. Designate their horizontal projections symmetrically about the horizontal axis of the circle.
Thus, the projections of the desired section are defined: on the plane P₂ - a straight line (points 1₂, 2₂ ... 10₂); on the plane P₁ - a circle (points 1₁, 2₁ ... 10₁).
By two, construct the natural size of the section of the given cylinder by the front-projecting plane Σ. To do this, use the method of projections.
Draw the plane P₄ parallel to the projection of the plane Σ₂. On this new x₂₄ axis, mark the point 1₀. Distances between points 1₂ - 2₂, 2₂ - 4₂, etc. from the frontal projection of the section, set aside on the x₂₄ axis, draw thin lines of projection connection perpendicular to the x₂₄ axis.
In this method, the P₄ plane is replaced by the P₁ plane, therefore, from the horizontal projection, transfer the dimensions from the axis to the points to the axis of the P₄ plane.
For example, on P₁ for points 2 and 3, this will be the distance from 2₁ and 3₁ to the axis (point A), etc.
Having postponed the indicated distances from the horizontal projection, you will get points 2₀, 3₀, 6₀, 7₀, 10₀, 11₀. Then, for greater accuracy of construction, the remaining, intermediate, points are determined.
By connecting all the points with a curved curve, you will obtain the desired natural size of the cross section of the cylinder by the front-projecting plane.
Sources:
- how to replace plane
Tip 3: How to find the area of the axial section of a truncated cone
To solve this problem, you need to remember what a truncated cone is and what properties it has. Be sure to draw. This will determine which geometric figure is a section. It is quite possible that after this the solution of the problem will no longer be difficult for you.
Instruction
A round cone is a body obtained by rotating a triangle around one of its legs. Straight lines coming from the top cones and intersecting its base are called generators. If all generators are equal, then the cone is straight. At the base of the round cones lies a circle. The perpendicular dropped to the base from the top is the height cones. At the round straight cones height coincides with its axis. The axis is a straight line connecting to the center of the base. If the horizontal cutting plane of the circular cones, then its upper base is a circle.
Since it is not specified in the condition of the problem, it is the cone that is given in this case, we can conclude that this is a straight truncated cone, the horizontal section of which is parallel to the base. Its axial section, i.e. vertical plane, which through the axis of a circular cones, is an isosceles trapezoid. All axial sections round straight cones are equal to each other. Therefore, to find square axial sections, it is required to find square trapezoid, the bases of which are the diameters of the bases of the truncated cones, and the sides are its generators. Truncated Height cones is also the height of the trapezoid.
The area of a trapezoid is determined by the formula: S = ½(a+b) h, where S is square trapezoid; a - the value of the lower base of the trapezoid; b - the value of its upper base; h - the height of the trapezoid.
Since the condition does not specify which ones are given, it is possible that the diameters of both bases of the truncated cones known: AD = d1 is the diameter of the lower base of the truncated cones;BC = d2 is the diameter of its upper base; EH = h1 - height cones.Thus, square axial sections truncated cones defined: S1 = ½ (d1+d2) h1
Sources:
- truncated cone area
The cylinder is a three-dimensional figure and consists of two equal bases, which are circles, and a lateral surface connecting lines bounding the bases. To calculate square cylinder, find the areas of all its surfaces and add them up.
The area of the lateral surface of a cylinder is equal to the area of a rectangle whose base is 2π r, and the height is equal to the height of the cylinder h, i.e. 2π rh.
The total surface of the cylinder will be: 2π r 2+2π rh= 2π r(r+ h).
The area of the lateral surface of the cylinder is taken sweep area its lateral surface.
Therefore, the area of the lateral surface of a right circular cylinder is equal to the area of the corresponding rectangle (Fig.) and is calculated by the formula
S b.c. = 2πRH, (1)
If we add the area of the two bases of the cylinder to the area of the lateral surface of the cylinder, we get the total surface area of the cylinder
S full \u003d 2πRH + 2πR 2 \u003d 2πR (H + R).
Straight cylinder volume
Theorem. The volume of a right cylinder is equal to the product of the area of its base and the height , i.e.where Q is the base area and H is the height of the cylinder.
Since the base area of the cylinder is Q, there are sequences of circumscribed and inscribed polygons with areas Q n and Q' n such that
\(\lim_(n \rightarrow \infty)\) Q n= \(\lim_(n \rightarrow \infty)\) Q' n= Q.
Let us construct sequences of prisms whose bases are the described and inscribed polygons considered above, and whose lateral edges are parallel to the generatrix of the given cylinder and have length H. These prisms are described and inscribed for the given cylinder. Their volumes are found by the formulas
V n= Q n H and V' n= Q' n H.
Hence,
V= \(\lim_(n \rightarrow \infty)\) Q n H = \(\lim_(n \rightarrow \infty)\) Q' n H = QH.
Consequence.
The volume of a right circular cylinder is calculated by the formula
V = π R 2 H
where R is the radius of the base and H is the height of the cylinder.
Since the base of a circular cylinder is a circle of radius R, then Q \u003d π R 2, and therefore
