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# Find the equation of the cylinder having its base the circle

find the equation of the cylinder having its base the circle x^2 + y^2 + (z-1)^2 = 4 , x-2y=1.

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### Cylinder - Wikipedia

Commonly the word cylinder is understood to refer to a finite section of a right circular cylinder having a ... cylinder of base ... cylinder, in a circle ...

### Surface, Curves and Equations | [email protected]

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How to express the standard form equation of a circle of a given radius. Practice problems with worked out solutions, ... Surface area of a Cylinder

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Formula Area of a Cylinder. How to Find a Cylinder's Area. Algebra ; Geometry ; Trigonometry; Calculus; Worksheets; Math Gifs; Teacher Tools ... Unit Circle Game

### Surface Area and Volume Formulas for Figures - ThoughtCo

Surface area and volume formulas for cylinders, ... Surface Area and Volume: Cylinder, Cone, Pyramid, Sphere, ... Area of a Circle Sector.

### Cylinder - Definition, Volume, Surface Area & Examples | Math ...

A circular cylinder having perpendicular base and height is known as a ... The normal to the plane of the circle through its center is the axis ... Cylinder Equation.

### Perimeter (Circumference) of Cylinder Calculator

Perimeter of Cylinder Calculator. ... A cylinder having its circular base with diameter 50 cm will have its circumference as ... Radius Of A Circumscribed Circle ...

## Suggested Questions And Answer :

### Find the equation of the cylinder having its base the circle

When z=1, x^2+y^2=4, which is a circle, centre at the origin and radius 2. x-2y=1 is a plane intercepting the x-y plane with y intercept=-1/2 and x intercept=1. The equation x^2+y^2+(z-1)^2=4 is a sphere, centre (0,0,1) and radius 2. The plane cuts through the sphere when (1+2y)^2+y^2=4; 1+4y+4y^2+y^2=4; 5y^2+4y-3=0, y=(-4±sqrt(16+60)/10=0.47178 ((-2+sqrt(19))/5) and -1.27178 ((-2-sqrt(19))/5). These values give x=1.94356 ((1+2sqrt(19))/5) and -1.54356 ((1-2sqrt(19))/5 respectively, by using the equation x=1+2y. So the points of intersection are (1.94356,0.47178) and (-1.54356,-1.27178). From these we can find the diameter of the cylinder: sqrt(76/25+304/25)=sqrt(380)/5=2sqrt(95)/5=3.9 approx. The radius is sqrt(95)/5. The sides of the cylinder are perpendicular to the diameter.  The slope of the diameter is the same as the slope of the plane x-2y=1, which is 1/2, so the slope of the perpendicular is -2. The equations of the sides of the cylinder where z=1 in the x-y plane are y=-2x+c where c is found by plugging in the intersection points: (-2+sqrt(19))/5=-2(1+2sqrt(19))/5+c; -2+sqrt(19)=-2-4sqrt(19)+5c. So c=sqrt(19) and y=sqrt(19)-2x as one side. Similarly, -2-sqrt(19)=-2+4sqrt(19)+5c, c=-sqrt(19) and y=-(sqrt(19)+2x) for the other side. Consider the view looking along the z axis. The circular cross-section of the cylinder will appear edge on, while the sides will have the slope -2 and will be spaced apart according to the value of z. When z=1±radius of cylinder=1±sqrt(95), the sides will appear to be as one line passing through the x-y plane's origin, the equation of the line being y=-2x when z=1±sqrt(95). The picture shows the view from z=1 looking at the x-y plane. The circle is the cross-section of the sphere and the line passing through (1,0) and (0,-1/2) is the edge of the plane x-2y=1. The diameter of the cylinder is constant and is shown by the two lines perpendicular to each end of the chord where the plane cuts the sphere. These lines represent the sides of the cylinder as they would appear at z=1. Parallel to them and passing through (0,0) is the single line that appears when z is at the extreme limits of the diameter of the cylinder. This line is also the central axis of the cylinder. (The vertical line at y=-2 is just a marker to show the the leftmost limit of the diameter of the sphere.) The general equation of a cylinder is the same as the 2-dimensional equation of a circle: x^2+y^2=a^2. This cylinder, radius a, has the z axis its central axis. x^2+z^2=a^2 is a cylinder with the y axis as its central axis. a=sqrt(95)/5 so a^2=95/25=19/5, making the equation of the cylinder 5x^2+5z^2=19. This is the same size as the cylinder in the problem, but with its central axis as the y axis. 5x^2+5(z-1)^2=19 is the equation of the cylinder with central axis passing through the centre of the sphere. If the cylinder is tilted so that its base coincides with the circle produced by the plane cutting through the original sphere, more calculations need to be made  to transform the coordinates. The picture shows the axial tilt of the cylinder. The central axis of the cylinder, y=-2x, bisects the chord on the line x-2y=1 when x-2(-2x)=1; 5x=1, x=1/5. So y=(x-1)/2=-2/5. If x-2y=1 represents the horizontal axis (we'll call X) and y=-2x represents the vertical axis (Y) we can see that, relative to these X-Y coordinates, the cylinder is upright. The  z value is unaffected by rotation. Take a point P(x,y) in the x-y plane. What are its coordinates in the X-Y plane? To find out we use geometry and trigonometry. The axial tilt of the X-Y axes is angle ø where tanø=1/2, so sinø=1/sqrt(5) and cosø=2/sqrt(5).  X=SP=QPcosø=2(x+y/2)/sqrt(5), Y=PRcosø=2(y-(x-1)/2)/sqrt(5). Z=z In the X-Z plane, 5X^2+5Z^2=19. This transforms to 4(x+y/2)^2+5z^2=19=(2x+y)^2+5z^2. In the X-Y plane the sides of the cylinder are the lines X=-a and X=a, where a^2=19/5. So X^2=19/5, which transforms to 4(x+y/2)^2 or (2x+y)^2=19/5; 2x+y=±sqrt(3.8), corresponding to the equation of two parallel, sloping line forming the sides of the cylinder.

### area under a curve at the right of a circle

Remember that the area under a curve has to be bounded, so limits need to be applied to define the boundaries. Usually it's the area between the curve and one axis or both. Occasionally it's the area between two curves or the area produced by the intersection of two curves. Sometimes, as in the case of a circle or ellipse which already encloses an area, it's the whole or part of an area inside the curve. The "formula" is based on the area of very thin rectangles, infinitely thin, in fact, which are laid side by side to fill the area. An integral is applied which sums the areas of the rectangles over the whole region specified to get the area of that region. The most common formula is integral(ydx) where y=f(x) defines the curve. Limits of x are then given to enclose the area, so that definite integration is applied. The best way to approach any problem involving finding areas related to a curve is to sketch the curve and draw the area that needs to be found. Imagine a large curve had been drawn on the ground and you had a roll of sticky tape. You work out where the area is and cut strips of tape and stick them down so that you fill the required area. You can only lay the strips side by side, no criss-crossing and no laying the tape in a different direction. You can only cut the strips into rectangles using a cut straight across, not at an angle. You end up with not quite filling, or slightly over-filling the space because the tape will overlap the curve slightly. But when you step back it will look like the area has been properly covered by tape. If you used narrower tape the area would be even better filled. That's the principle on which the integration is based, because the area of each rectangle is the length of the strip y times the width we call dx (the width of the strips of tape). The area outside a circle must be bounded. You need the equation of the circle so that you can relate x and y. For example, the general equation of a circle is (x-h)^2/r^2 + (y-k)^2/r^2=1, where r is the radius and (h,k) is the centre. So y=k+sqrt(r^2-(x-h)^2) is half the circle, because the other half is k-sqrt(r^2-(x-h)^2).  If you need the area between the circle and the x axis between a and b, you need the lower part of the curve given by the second expression; if you need the upper part of the curve you need the first expression. If the circle is coloured red and the outside of the circle is blue, the area between the lower circle and the x axis will be entirely blue, whereas the area between the upper circle and the x axis will be red and blue. Get the picture? Once you've decided what you want, you compose the integral: integral(ydx)=integral((k-sqrt(r^2-(x-h)^2))dx) for b Read More: ...

### How do you find the area of the base of a cylinder with a radius that is 4?

Problem: How do you find the area of the base of a cylinder with a radius that is 4? I just need to figure it out or learn it somehow. By stating "radius," you are saying that the base is a circle. The area of a circle is A = pi * r^2 A = pi * r^2 A = 3.1415 * 4^2 A = 3.1415 * 16 A = 50.264 square units, whatever the 4 is

### how to calculate volume of a cylinder when the surface of the contence has an elliptical surface(not flat and not a cone)?

For an elliptical cylinder we use the formula for the area of an ellipse instead of a circle before we multiply by the height. The area of an ellipse is (pi)ab where ab replaces r^2 for a circle. The values of a and b are the lengths of the major and minor axes of the ellipse. When a=b we have a circle. So the volume of the cylinder is (pi)abh, where h is the height of the elliptical cylinder. The general rule for finding the volume of an object with a constant cross-section is to multiply the height (length) of the object by the area of its base, which could be circular, triangular, rectangular, square, octagonal, star-shaped, a parallelogram, a mixture, or whatever.

### find the surface area of a cylinder. height-4 width-5 base-6

First take the top and bottom discs, which each have an area of (pi)r^2. The surface area of the curved side is 2(pi)rh  (circumference times height) where r=radius and h=height. So the total surface area is 2(pi)r(r+h). But which value do we use for the width? If we use the base, r=3 because the base=diameter=2r. The surface area is 2*(pi)*3*7=42(pi)=131.95 cubic units. If we use the width, r=2.5 and the surface area is 2*(pi)*2.5*6.5=32.5(pi)=102.10. If the width is the thickness of a hollow cylinder we take the difference between the inner and outer top and bottom circles=2(pi)(3^2-2.5^2)=5.5(pi). The outer surface area is 2(pi)*3*4=24(pi) and the inner surface area is 2(pi)*2.5*4=20(pi). These curved surface areas have to be added together=44(pi). Altogether then the total surface area is (5.5+44)(pi)=49.5(pi)=155.51 cubic units.

### In a triangle ABC, line joining orthocenter and the circumscenter is parallel to AC, then find tanA *tanC.

Draw the base of the triangle AC and mark its midpoint, D. Draw a perpendicular from D. Somewhere along the perpendicular will be the circumcentre, that is, the centre of the circle that circumscribes the triangle ABC.  The position of vertex B has to be such that the perpendicular bisector of AB meets the perpendicular bisector of AC at X, and the perpendicular from vertex C on to AB meets the perpendicular from vertex B on to AC at Y, such that XY is parallel to AC. This is another way of framing the question. No other construction lines are required because the missing perpendiculars are superfluous to solving the problem. Represent the problem graphically. Plot the points A(0,0), B(p,q), C(t,0) where p and q are to be determined and t is an arbitrary constant t=AC. Midpoint of AC is N(t/2,0); midpoint of AB is M(p/2,q/2). AB is a segment of the line y=qx/p. The equation of the bisector of AB: -p/q is its gradient, so y=-px/q+c, where c is found by plugging in M: q/2=-p^2/2q+c, c=q/2+p^2/2q and the perpendicular bisector is a segment of y=q/2+p^2/2q-px/q. Therefore, the coords of X are where this line meets the perpendicular bisector of AC, which is a segment of the line x=t/2. The intersection is X(t/2,q/2+p^2/2q-pt/2q). Now we need to find Y, which must lie on the line x=p, because this is the perpendicular from vertex B on to AC. The perpendicular from vertex C on to AB is parallel to the perpendicular bisector of AB so has the same gradient: -p/q. The equation of the perpendicular from C is y=-px/q+k, where k is found by plugging in C(t,0): 0=-pt/q+k, k=pt/q and y=-px/q+pt/q=p(t-x)/q. The point Y is therefore Y(p,p(t-p)/q). XY is parallel to AC, which means their y coord is the same: q/2+p^2/2q-pt/2q=p(t-p)/q; q^2+p^2-pt=2p(t-p); q^2+3p^2=3pt; q^2=3p(t-p). Also, tanA=q/p and tanC=q/(t-p), so tanA*tanC=q^2/(pt-p^2)=3p(t-p)/(p(t-p))=3. Note that when tanA=tanB=sqrt(3), B is (1/2,sqrt(3)/2), the triangle is equilateral and XY=0 because the circumcentre and orthocentre coincide.

### How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.

### solid that remain after drilling a hole of radius b through the center of a sphere of radius R (b<R)

Consider a circle with equation x^2+y^2=R^2. Consider a circle with equation x^2+y^2=b^2. The latter represents the cross-section of the hole drilled through the centre of the cross-section of the sphere represented by the circle with the first equation. The graph looks like a view of the sphere with the drilled hole from its top, looking down the hole. Now, consider a concentric circle with radius between b and R. That radius can be represented as a value x along the x axis, which is the radius of a cylinder concentric with the drilled hole. The circumference of the circle with radius x is 2(pi)x and the surface area of the cylinder with the same radius is 2(pi)xh where h is the length of the cylinder and the cylinder itself, and this length is perpendicular to the xy plane, "into the paper", so to speak. Now imagine a side view of the sphere and drilled hole. In this view, the circle is still represented in two dimensions by x^2+y^2=R^2 and the hole is a line segment on each side of the y axis at x=b and x=-b. The point we marked as x between b and R is now somewhere on the circumference of the side view of the sphere and its y value in this side view is h, so h=2sqrt(R^2-x^2). The reason for 2 is that h extends from x at its entry point through the x axis and to the same distance below the x axis at the exit point. If we now unfold the cylinder into rectangle with length h and width equal to the circumference of the cylinder's circular end we have a rectangle equal in area to the surface area of the cylinder. Give this an infinitesimal thickness dx and we have the volume of a very thin cylindrical shell that runs through the sphere. This volume is 2(pi)xhdx=4(pi)x*sqrt(R^2-x^2)dx. If we integrate this between the limits b and R we should end up with the volume of material remaining after drilling a hole radius b through the centre of a sphere radius R. Let u=R^2-x^2, then du/dx=-2x or xdx=-du/2. The integrand becomes -2(pi)u^(1/2)du which when integrated becomes -2(pi)u^(3/2)*(2/3)=-4/3(pi)(R^2-x^2)^(3/2) between the limits x=b to R. This evaluates to 4/3(pi)(R^2-b^2)^(3/2). When b=0 this is 4/3(pi)R^3, the volume of the sphere, and when b=R it's zero.