We didn't use terms like by shells and by cylinders in lecture so I'm trying to follow along, but since if I'm doing by cylinders would I then require the constant to be 2pi?
Homework Statement
Find the volume of the solid of revolution obtained by rotating the area bounded by the curves about the line indicated.
y=x2-2, y=0 about y=-1. Need only consider part above y=-1
Homework Equations
V=∏a∫b[f(x)]2dx
The Attempt at a Solution
I'm mainly unsure of...
I'm thinking I should only consider the integral of one half and then double it, as it is symmetrical.
I know the total range of x is from 0 to 1. I'll have to integrate this in two parts as the upper and lower function alternate.
The first would be the integral from 0 to the meeting point of...
So the range of y above the line is from a(the line) to 1
The value of x is -√1-y2, if it's to the left of the origin at the midpoint.
If I integrate in terms of x, say, I have the integral from 0 to some point of (equation of circle - a) + the integral from that same point to 1 of...
Homework Statement
The figure shows a semicircle with radius 1, horizontal diameter , and tangent lines at
and . At what height above the diameter should the horizontal line be placed so as to minimize
the shaded area
http://imgur.com/grrCqWF
Homework Equations
The equation of a...
So should the x inside the integral be replaced with an f(x)? That's the equation the prof gave in class, however in examples he switched back and forth between using f(x) and x. I'll ask about it next class.
Thanks!
Homework Statement
Obtain the surface area when the curve y=ex, 0≤x≤1, is rotated about the x-axis
Homework Equations
Surface Area = 2∏a∫b x√(1+(dy/dx)2)dx
The Attempt at a Solution
I started with the the equation, Surface Area = 2∏0∫1 x√(1+e2x)dx. However, whichever way I try to...
Homework Statement
Find d2/dx2 0∫x (1∫sint√(1+u^4)du)dtHomework Equations
The Attempt at a Solution
Initially I treated this problem as the second derivative of a double integral and thus quickly found myself at the result cosx√1+sin4x, by the fundamental theorem of calculus. However I...