Let R be the region bounded by y=6sin((pi/2)x), y=6(x-2)^2, y=3x+3 containing the point (2,6). Find the area of R, find the volume of R rotated about the x-axis, find the volume of R rotated about the y-axis, and Suppose R is the base of a shape in which cross-sections perpendicular to the x-axis are squares. Find the volume of this shape.
Major arithmetic mess.
After making quick sketch, it was easy to see that all 3 graphs intersect at (1,6)
the sine curve and the parabola also intersect at (2,0)
and the straight line intersects the parabola at (3.5 , 13.5)
The straight line also intersects the curve at appr .58 and again in the third quadrant, but those two intersection points do not lie in our region
Shade in the region containing the point (2,6).
(This point does not enter any calculations, it was merely given to identify the region we want)
We will have to split our region into two parts, draw a vertical line from (2.0) to the line y=3x+3
Rotation around x-axis:
Part 1. from x = 1 to 2 ...
outer radius = 3x+3
inner radius = 6sin(πx/2)
Volume = π∫(3x+3)^2 - 36sin^2 (πx/2) dx from x=1 to x=2
integral of (3x+3)^2 is not too bad , can be done mentally ..
= (1/3)(1/3)(3x+3)^3 = (1/9)(3x+3)^3
the integral of 6 sin^2 (πx/2) is a bit messier
use cos 2A = 1 - 2sin^2 A
cos πx = 1 - 2sin^2 (πx/2)
3cos πx = 3 - 6sin^2 (πx/2)
6sin^2 (πx/2) = 3 - 3cos πx
and the integral of 6sin^2 (πx/2) is 3x + (3/π)cos πx
so the volume of the first part is
π[ (1/9)(3x+3)^3 + 3x + (3/π)cos πx ] from 1 to 2
= π( 81 + 6 + 3/π - (24 + 3 - 3/π) )
= π(60)
= 60π
and now for the 2nd part:
Volume = π∫(3x+3)^2 - 36(x-2)^4 dx from x = 2 to 3.5
= π[ (1/9)(3x+3)^3 - (36/5)(x-2)^5 ] form 2 to 3.5
= appr 137.7π
so volume when rotated about x-axis is
appr 197.7π or 621.1
arggghhh, where did you get this question??
rotating about the y-axis is even more fun.
from y = 6 sin (πx/2) we will need x^2 !!!!!!
πx/2 = sin^-1 (y/6)
x = (2/π) sin^-1 (y/6) )
we now have to square that, and then integrate ....
I YIELD!
It's a homework problem from my Calculus II class...I tried to attempt and failed miserably. Thank you so much for your help. This sheds a little light so hopefully the rest aren't as bad.
To find the area of region R:
1. We need to find the points at which the curves intersect.
Set y = 6sin((π/2)x) equal to y = 6(x-2)^2:
6sin((π/2)x) = 6(x-2)^2
Divide both sides by 6:
sin((π/2)x) = (x-2)^2
2. From the graph, we can see that the curves intersect at x = 0 and x = 4.
So, the region of interest lies between x = 0 and x = 4.
3. To find the area, we need to integrate the difference between the two curves over this interval.
A = ∫[0,4] (6(x-2)^2 - 6sin((π/2)x)) dx
4. Evaluating this integral will give us the area of the region R.
To find the volume of R rotated about the x-axis:
1. We need to use the washer method.
The formula for the volume of a solid of revolution using the washer method is:
V = π∫[a,b] (R(x)^2 - r(x)^2) dx
2. In this case, R(x) is the outer radius and r(x) is the inner radius.
R(x) = 6(x-2)^2 and r(x) = 3x+3
3. The bounds for the integral are still from x = 0 to x = 4.
4. Plug in the values and evaluate the integral to find the volume.
To find the volume of R rotated about the y-axis:
1. We need to use the shell method.
The formula for the volume of a solid of revolution using the shell method is:
V = 2π∫[c,d] y(x) * h(x) dx
2. In this case, y(x) is the height of the cylinder and h(x) is the circumference of the cylinder at a given height.
y(x) = 6sin((π/2)x) - 6(x-2)^2
h(x) = 2πx
3. The bounds for the integral will depend on the range of y values that correspond to region R.
We need to find the values of x that satisfy the inequalities y = 6sin((π/2)x) > 6(x-2)^2 and y = 6sin((π/2)x) > 3x+3.
4. Once we find the bounds for the integral, plug in the values and evaluate the integral to find the volume.
To find the volume when R is the base of a shape with perpendicular square cross-sections:
1. Each square cross-section will have side length equal to the height of the corresponding point on the curve.
2. The height of each point on the curve is given by y = 6sin((π/2)x) - 6(x-2)^2.
3. To find the volume, we need to integrate the area of each square cross-section over the interval [0,4].
4. The formula for volume in this case will be V = ∫[0,4] (y(x))^2 dx.
5. Plug in the values and evaluate the integral to find the volume.
To find the area of the region R bounded by the given curves, we need to find the points of intersection first.
1. Find the points of intersection between y = 6sin((π/2)x) and y = 6(x-2)^2:
Set the two equations equal to each other: 6sin((π/2)x) = 6(x-2)^2
Simplify the equation: sin((π/2)x) = (x-2)^2
The only point of intersection is x = 2, since sin((π/2)x) and (x-2)^2 are never equal for any other value of x.
Plug this value into either equation to find the corresponding y-coordinate:
y = 6sin((π/2) * 2) = 6sin(π) = 0
Therefore, the point of intersection is (2, 0).
2. Find the points of intersection between y = 6sin((π/2)x) and y = 3x + 3:
Set the two equations equal to each other: 6sin((π/2)x) = 3x + 3
Simplify the equation: sin((π/2)x) = (1/2)x + 1/2
There are two points of intersection which can be found by solving this equation:
x = 0: sin(0) = 0 + 1/2, which is not true.
x = 4: sin((π/2) * 4) = (1/2) * 4 + 1/2 = 3
Plug this value into either equation to find the corresponding y-coordinate:
y = 6sin((π/2) * 4) = 6sin(2π) = 0
Therefore, the point of intersection is (4, 0).
Now we can proceed to find the area of region R using the definite integral.
3. Area of R: The area of region R can be calculated by integrating the difference between the two curves with respect to x, from x = 2 to x = 4:
A = ∫[2 to 4] (6sin((π/2)x) - 6(x-2)^2) dx
To find the volume of R rotated about the x-axis, we will use the disk method.
4. Volume of R rotated about the x-axis: The volume can be calculated by integrating the area of the circular cross-sections perpendicular to the x-axis, from x = 2 to x = 4:
V = ∫[2 to 4] π[(6sin((π/2)x))^2 - (6(x-2)^2)^2] dx
To find the volume of R rotated about the y-axis, we will use the shell method.
5. Volume of R rotated about the y-axis: The volume can be calculated by integrating the areas of the cylindrical shells perpendicular to the y-axis, from y = 0 to y = 6:
V = ∫[0 to 6] 2πx[(6sin((π/2)x) - (3x + 3)] dy
Lastly, we need to find the volume of the shape where the cross-sections perpendicular to the x-axis are squares.
6. Volume of shape with square cross-sections: The volume can be calculated by integrating the areas of the square cross-sections perpendicular to the x-axis, from x = 2 to x = 4:
V = ∫[2 to 4] ((6sin((π/2)x))^2 - (6(x-2)^2)^2) dx
By evaluating these integrals, you can find the exact values of the area and volumes for region R and the shape with square cross-sections.