Surface area of sphere

A sphere is a three-dimensional geometric object which is perfectly in round shape. Mathematically, the sphere is a set of points equidistant from a single point. That common point of the sphere is called the center and distance between common point and any other point on the sphere is called the radius. Sphere Calculator: Finding the sphere volume, circumference, radius, and surface area becomes quite easy with our online calculator. This article gives clear information regarding the simple formulas, step by step process to calculate the unknown measures of a sphere using the known details. Apart from this, you will also find the solved examples User-friendly geometry tool Sphere Calculator computes the unknown parameters of a sphere effortlessly by taking the required details. You just need to provide known data values as input and press the calculate button to check the accurate output within no time. Below mentioned are the basic and straightforward formulas to compute the sphere volume, circumference, and surface area by using the radius. • Volume of a Sphere Formula: • Volume V = (4/3)πr³ • Circumference of a Sphere Formula: • Circumference C = 2πr • Surface Area of a Sphere Formula: • Surface Area A = 4πr² Sphere Calculations: This section gives the sphere formulas in terms of volume, circumference, and surface area. • If sphere circumference is known, then • Volume V = (C³ / 6π²) • Radius r = C / 2π • Surface Area A = C² / π • If the surface area of a sphere is giv...

Spheres - Higher tier only A sphere is a perfectly round solid figure. All points on the surface of the shape are the same distance away from the centre – we call this distance the radius. The formula for the volume and surface area of a sphere will be given to you in the exam, so you will not need to memorise these. \[\text\] The diameter is the distance from one point on the surface to another, through the centre. If you are given the diameter you must divide by 2 to find the radius before you can calculate the volume or surface area. Example A spherical fish tank of diameter 40 cm is half full of water. 1. Calculate the volume of the water 2. The water is transferred into a new spherical tank so that the water fills the tank completely. Calculate the surface area of the new tank Solution 1. Calculate the volume of the tank: Diameter = 40 cm so the radius is 40 ÷ 2 = 20 cm Substitute this into the formula for the volume of a sphere: Volume = \(\frac \times \pi \times\) 15.87401052 2 = 3166.526972 cm 2 Surface area = 3,166.53 cm 2 (to two decimal places) Question A rubber band ball has a radius of 6 cm. I add some more rubber bands and the volume increases by 100 cm 3 . How much has the radius increased by? Give your answer to two decimal places. Reveal answer down Old volume = \(\frac\) = 6.213370741 cm = 6.21 cm (to 2 decimal places) The radius has increased by 0.21 cm Question A lipstick container has a diameter of 12 mm and a height of 52 mm. It consists of a hemisphe...

\( \newcommand\,\Delta A_i.\] Once again take a limit as all of the \(\Delta x_i\) and \(\Delta y_i\) shrink to 0; this leads to a double integral. Definition 105 Surface Area Let \(z=f(x,y)\) where \(f_x\) and \(f_y\) are continuous over a closed, bounded region \(R\). The surface area \(S\) over \(R\) is \[\begin\] Note: as done before, we think of "\(\displaystyle \iint_R\ dS\)'' as meaning "sum up lots of little surface areas over \(R\).'' The concept of surface area is defined here, for while we already have a notion of the area of a region in the plane, we did not yet have a solid grasp of what "the area of a surface in space'' means. We test this definition by using it to compute surface areas of known surfaces. We start with a triangle. Example \(\PageIndex.\] We affirm the validity of our formula. It is "common knowledge'' that the surface area of a sphere of radius \(r\) is \(4\pi r^2\). We confirm this in the following example, which involves using our formula with polar coordinates. Example \(\PageIndex\) fails the requirements of Definition 105, as \(f_x\) and \(f_y\) are not continuous on the boundary of the circle \(x^2+y^2=a^2\). The computation of the surface area is still valid. The definition makes stronger requirements than necessary in part to avoid the use of improper integration, as when \(f_x\) and/or \(f_y\) are not continuous, the resulting improper integral may not converge. Since the improper integral does converge in this example, the surface a...