1. ## Calculus I

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.

2. Originally Posted by Caycee7
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?
Here's the first part of question 1, one attachment is the diagram and one is the solution

3. Originally Posted by Caycee7
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 the region R above is rotated acout the line y=2?
Here's the second part of question 1

4. Originally Posted by Caycee7

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?

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

5. Originally Posted by Caycee7
2.Consider the region S bounded by the y-axis, the line y= cosx and the line y=sinx.

What is the volume of the region generated by rotating S about he x-axis?
Here's the second part of 2

6. Originally Posted by Caycee7
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 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

7. ## more questions....

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.

8. Originally Posted by Caycee7
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

9. This was my firtst attempt

10. Originally Posted by Caycee7
This was my firtst attempt

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

11. ah. yeah that makes sense...
thanks once again.