There are a few Calc problems that have me stumped, can anyone help me out!?
1 .Consider the region R bounded by the curve y=sinx and the x-axis on the intercal [o, pi/2]. What is the volume of the solid generated when R is rotated about the x-axis?
What is the volume of the solid generated when the region R above is rotated acout the line y=2?
2.Consider the region S bounded by the y-axis, the line y= cosx and the line y=sinx. What is the area of S?
What is the volume of the region generated by rotating S about he x-axis?
3. Consider the region Q bounded by the x-axis, the y-axis and the line y= squareroot of 16-x^2. What is the volume of the solid generated by rotating the region Q about the y-axis.
Here's the solution to the first part of question 2, again, one attachment is a diagram the other is the solution
I assume you are talking about the region in the first quadrant, the region in the second and third quad would be too complicated for what i think you are doing. you should specify which you mean though
Here's question 3. i attached both the diagram and solution
Again i assumed you're talking about the region in the 1st quadrant, but in this case that doesn't matter. the graph is symmetric about the y-axis, so if i assumed it was the 2nd quad, we'd get the same answer.
EDIT: I made a mistake for the diagram. it shows that we are rotating about the x-axis when in fact we are rotating about the y-axis. you can change it
Ha. thanks for pointing out where it is, I know wasnt there earlier, or perhaps my vision is worse than previously suspected...
Well actually on the subject of understanding, if you dont mind:
For part 1a, i followed you for the set up and up untill the part when I think you used the pathagorean identiy to make sinx^2 into cosx+1.....
I think that it is suppose to be 1- cosx^2. and I am not sure how you pulled the pi/2 out of the integrals and got rid of the squared and made it 2x.
well, you were correct in that i had the wrong thing. however, I did not use the pythagorean identity, i used the half-angle identity.
You thought i used this: sin^2(x) = 1 - cos^2(x)
In fact, i meant to use this: sin^2(x) = (1 - cos(2x))/2
that's where i got the pi/2 from. i pulled out the 1/2 in front.
changing sin^2(x) to 1 - cos^2(x) would not help, since integrating the squared trig function was the problem in the first place, i had to get rid of the square, so i used the identity above. however, the plus sign should be a minus sign, make the necessary corrections
first off, the square is not on the x, it is on the sin(x), so write sin^2(x) or (sin(x))^2
secondly, you cannot integrate squared trig functions like that, the integral of sin^2(x) IS NOT -cos^2(x). you can't just forget the square and continue as normal, you must get rid of it.
also if your method was right, you would end up with +pi