Let R be the region bounded by the following curves. Use the disk (washer) method to find the volume of the solid generated when R is revolved about the y-axis.
y=x, y=3x, y=6
Set up the integral that gives the volume of the solid. Use increasing limits of integration.
Limits of the integral is 0 to 6. I need help with the integrand. I thought it would be π(y/3)^2-(y^2), however that was incorrect.
To find the volume of the solid generated when R is revolved about the y-axis using the disk (washer) method, we need to integrate the area of the cross-sections along the y-axis.
The region R is bounded by the curves y = x, y = 3x, and y = 6. To find the correct integrand, we need to break down the region into two parts due to the intersection point of y = x and y = 3x.
Let's determine the limits of integration for each part:
1. For the region below y = 3x:
The outer radius of the washer is determined by the curve y = 6 and the inner radius is determined by the curve y = x. Thus, the area of the cross-section is given by πR^2 - πr^2, where R represents the outer radius and r represents the inner radius.
The outer radius (R) is y = 6, and the inner radius (r) is y = x.
2. For the region above y = 3x:
Here, the outer radius is still y = 6, but the inner radius changes. Now, the inner radius is determined by the curve y = 3x.
Now, let's set up the integral for each part:
1. Integrating for the region below y = 3x:
First, determine the limits of integration. Since y = 3x and y = 6 intersect at x = 2, we have two parts for this region: from y = 0 to y = 6 and from y = 0 to y = 3.
Integrating over the interval from y = 0 to y = 3:
Volume_1 = ∫[0 to 3] (π * (6^2) - π * (x^2)) dx
2. Integrating for the region above y = 3x:
The limits of integration for this part are from y = 3 to y = 6.
Volume_2 = ∫[3 to 6] (π * (6^2) - π * (3x^2)) dx
Now, the total volume of the solid generated when R is revolved about the y-axis is the sum of Volume_1 and Volume_2:
Total Volume = Volume_1 + Volume_2
Please note that you might need to evaluate the integrals using the Fundamental Theorem of Calculus or any other applicable integration techniques.
To find the volume of the solid generated when R is revolved about the y-axis using the disk (washer) method, we need to integrate the area of the cross-sections perpendicular to the y-axis.
Let's break down the problem into two parts based on the given curves.
1. The region bounded by y = x and y = 3x:
First, we need to determine the limits of integration for this part of the region. The curves intersect at y = 0 and y = 6, so the limits of integration for this part will be from y = 0 to y = 6.
Now, let's consider an arbitrary height y within this range. At height y, the outer radius of the washer is given by the distance from the y-axis to the curve y = 3x, which is equal to x = y/3. The inner radius is given by the distance from the y-axis to the curve y = x, which is equal to x = y. Therefore, the radius of the washer is (y/3) - y = -2y/3.
The area of each washer is given by π((outer radius)^2 - (inner radius)^2). Substituting the values:
Area = π[((y/3) - y)^2 - (y^2)]
= π[(y^2/9) - (2y^2/3) + (4y^2/9) - (y^2)]
= π[y^2/9 - 2y^2/3 + 4y^2/9 - y^2]
= π[-(4y^2/9) + 2y^2/3]
2. The region bounded by y = 3x and y = 6:
Similarly, for this part, the limits of integration will be from y = 0 to y = 6.
At height y, the outer radius of the washer is given by x = y/3, and the inner radius is given by x = 6/3 = 2. Therefore, the radius of the washer is (y/3) - 2.
The area of each washer is given by π[((y/3) - 2)^2 - (y^2)]. Substituting the values:
Area = π[((y/3) - 2)^2 - y^2]
= π[(y^2/9) - (4y/3) + 4 - y^2]
= π[-(8y/3) + (4 - 8/9)y^2]
Now, we need to set up the integral to find the total volume V:
V = ∫[0, 6] [-(4y^2/9) + 2y^2/3] dy + ∫[0, 6] [-(8y/3) + (4 - 8/9)y^2] dy
Evaluating this integral will give us the volume of the solid generated when R is revolved about the y-axis.
each disc has thickness dy, so stack them up. Since we're integrating on y, express each function as x = f(y)
v = ∫[0,6] π(R^2-r^2) dy
where R=y and r=y/3
v = ∫[0,6] π(y^2-(y/3)^2) dy
check, using cylinders of thickness dx. You have to break the region into two parts, since the upper boundary changes at (2,6)
v = (∫[0,2] 2πrh dx) + (∫[2,6] 2πrh dx)
where (r=x and h=3x-x) and (r=x and h=6-x)
v = (∫[0,2] 2πx*2x dx) + (∫[2,6] 2πx(6-x) dx)
Which expressions can be added to find the volume of the solid figure?
Select all that apply.
A solid shape is made up of 2 attached rectangular prisms. First rectangular prism has a length of 3 m, height of 1 m, and width of 4 m. Second rectangular has a length of 3 m, height of 7 m, and width of 4 m
The total length of the shape is 6 meters.
A.4×3×7 and 1×3×4
B.4×6×7 and 1×3×3
C.1×4×7 and 4×3×6
D.4×6×1 and 6×4×3
E.4×6×3 and 7×3×3
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