Science:Math Exam Resources/Courses/MATH103/April 2012/Question 03 (a)
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Question 03 (a) 

Consider the area A between a circle of radius 1 in the first quadrant and ƒ(x)=x. Find the volume of the solid obtained by rotating A about the xaxis. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

Because the expression is a function for x and is to be rotated about the xaxis, we should use the disc method. When using the disc method, we want to evaluate the integral where a and b are the smallest and largest values of x in the region, f(x) is the largest yvalue in the region at x (which will be the outer radius of the disc) , and g(x) is the smallest yvalue in the region at x (which will be the inner radius of the disc). For this region, we have that . The rightmost xvalue is the point of intersection of the circle and the line . Try to find the value of b and evaluate the integral. 
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Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. The equation of a circle with radius 1 is x^{2}+y^{2}=1 or, once we solve for , (which gives the upper half of the circle). Because we are rotating around the xaxis, we will be integrating between two points on the xaxis. The left endpoint is clearly 0. The right endpoint is a little trickier: it will be the point where and intersect. This point of intersection is found by setting these two curves equal and solving:
Because we are in the first quadrant, the intersection point must be positive. Thus we will be integrating between and If we use the Disc Method, the outside radius of the disc will be the arc of the circle, given by and the inside radius will be given by the line . Plugging these, and our endpoints, into the integral, we get: 