1.
Find the volume of the solid obtained by rotating the region bounded by x = y^2 and x = 3y about the y-axis.
2. Find the volume of the solid obtained by rotating the region in the first quadrant bounded by y = x^2 and y = 9 about the y-axis.
I am taking an online class and the teachers notes are not helping. To anyone who takes the time to help me out, thank you so much.
Find the volume of the solid obtained by rotating the region bounded by x = y^2 and x = 3y about the y-axis.
2. Find the volume of the solid obtained by rotating the region in the first quadrant bounded by y = x^2 and y = 9 about the y-axis.
I am taking an online class and the teachers notes are not helping. To anyone who takes the time to help me out, thank you so much.
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1) To rotate the region bound by x = y^2 and x = 3y about the y axis, we will use the washer method. The washer method states that the integral for the volume is:
V(y) = π∫(Ro(y)^2 - Ri(y)^2)dy
Ro(y) is the radius for the outside surface of the washer. This is the function of y that is the farthest from the axis of rotation. For this problem that is x = 3y
Ri(y) is the radius for the inside surface of the washer. This is the function of y that is the closest to the axis of rotation. For this problem that is x = y^2
Our integral is:
V(y) = π∫((3y)^2 - (y^2)^2)dy
V(y) = π∫(9y^2 - y^4)dy
V(y) = π(3y^3 - (1/5)y^5)
We need to identify the interval over which we are integrating over. This can be found by putting both curves (x = y^2 and x = 3y) equal to each other and solving for y. Doing this we find:
y^2 = 3y
y^2 - 3y = 0
y(y - 3) = 0
Therefore our interval is from y = 0 to y = 3. Our volume is:
V = V(3) - V(0)
V = π(3*3^3 - (1/5)*3^5) - π(3*0^3 - (1/5)*0^5)
V = π(3*27 - (1/5)*243) - π(0 - 0)
V = π(81 - 243/5) - π(0)
V = π(405/5 - 243/5) - 0
V = π(162/5) - 0
V = 162π/5
2) To rotate the region bound by y = x^2 and y = 9 in the first quadrant about the y axis, we will use the shell method. The shell method states that the volume integral is:
V(x) = 2π∫(x)(p(x) - h(x))dx
Where
(x) represents the radius that we are rotating about. Since we are rotating about the y axis, we can leave this as x.
p(x) is the function of x that has the largest value over the interval of the integral. From drawing out the problem we see that p(x) = 9
h(x) is the function of x that has the smallest value over the interval of the integral. h(x) = x^2
V(x) = 2π∫(x)(p(x) - h(x))dx
V(x) = 2π∫(x)(9 - x^2)dx
V(x) = 2π∫(9x - x^3)dx
V(x) = 2π((9/2)x^2 - (1/4)x^4)
We need to determine the interval over which we are integrating. This can be found by putting both curves (y = x^2 and y = 9) equal to each other and solving.
x^2 = 9
x = 3 and x = -3
Since we are in the first quadrant, we will be using x = 0 and x = 3 as our interval. Therefore:
V = V(3) - V(0)
V = 2π((9/2)*3^2 - (1/4)*3^4) - 2π((9/2)*0^2 - (1/4)*0^4)
V = 2π((9/2)*9 - (1/4)*81) - 2π(0 - 0)
V = 2π(81/2 - 81/4) - 2π(0)
V = 2π(162/4 - 81/4) - 0
V = 2π(81/4)
V = 81π/2
V(y) = π∫(Ro(y)^2 - Ri(y)^2)dy
Ro(y) is the radius for the outside surface of the washer. This is the function of y that is the farthest from the axis of rotation. For this problem that is x = 3y
Ri(y) is the radius for the inside surface of the washer. This is the function of y that is the closest to the axis of rotation. For this problem that is x = y^2
Our integral is:
V(y) = π∫((3y)^2 - (y^2)^2)dy
V(y) = π∫(9y^2 - y^4)dy
V(y) = π(3y^3 - (1/5)y^5)
We need to identify the interval over which we are integrating over. This can be found by putting both curves (x = y^2 and x = 3y) equal to each other and solving for y. Doing this we find:
y^2 = 3y
y^2 - 3y = 0
y(y - 3) = 0
Therefore our interval is from y = 0 to y = 3. Our volume is:
V = V(3) - V(0)
V = π(3*3^3 - (1/5)*3^5) - π(3*0^3 - (1/5)*0^5)
V = π(3*27 - (1/5)*243) - π(0 - 0)
V = π(81 - 243/5) - π(0)
V = π(405/5 - 243/5) - 0
V = π(162/5) - 0
V = 162π/5
2) To rotate the region bound by y = x^2 and y = 9 in the first quadrant about the y axis, we will use the shell method. The shell method states that the volume integral is:
V(x) = 2π∫(x)(p(x) - h(x))dx
Where
(x) represents the radius that we are rotating about. Since we are rotating about the y axis, we can leave this as x.
p(x) is the function of x that has the largest value over the interval of the integral. From drawing out the problem we see that p(x) = 9
h(x) is the function of x that has the smallest value over the interval of the integral. h(x) = x^2
V(x) = 2π∫(x)(p(x) - h(x))dx
V(x) = 2π∫(x)(9 - x^2)dx
V(x) = 2π∫(9x - x^3)dx
V(x) = 2π((9/2)x^2 - (1/4)x^4)
We need to determine the interval over which we are integrating. This can be found by putting both curves (y = x^2 and y = 9) equal to each other and solving.
x^2 = 9
x = 3 and x = -3
Since we are in the first quadrant, we will be using x = 0 and x = 3 as our interval. Therefore:
V = V(3) - V(0)
V = 2π((9/2)*3^2 - (1/4)*3^4) - 2π((9/2)*0^2 - (1/4)*0^4)
V = 2π((9/2)*9 - (1/4)*81) - 2π(0 - 0)
V = 2π(81/2 - 81/4) - 2π(0)
V = 2π(162/4 - 81/4) - 0
V = 2π(81/4)
V = 81π/2