Undergrad Should we consider negative axis when finding the volume?

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The discussion centers on finding the volume of a solid formed by rotating the region bounded by the curves y² = x and x = 2y around the y-axis. A participant initially multiplied the integral by 2, assuming the limits included negative values, but was corrected that only the area above the x-axis was relevant for this problem. The curves intersect at (0, 0) and (4, 2), and the region to be revolved is entirely in the first quadrant. The confusion arose from plotting the curve y² = x, leading to a misunderstanding about including negative values. Ultimately, the correct approach focuses solely on the area above the x-axis for calculating the volume.
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Question:
Find the volume of the solid obtained by rotating the region
bounded by the given curves about the specified line. Sketch the
region, the solid, and a typical disk or washer.
y2 = x; x = 2y; about the y-axis

Solution: http://www.slader.com/textbook/9780538497909-stewart-calculus-early-transcendentals-7th-edition/438/exercises/9/

Now, when I solved the question, I multiplies the entire integral by 2 since the limits I took were from -2 to 2. However, in the answer, they have considered only the values above the x-axis and not below. Why is it so? It doesn't specify in the question that we need to find the area of only that region which lies above the x-axis.
 
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Phys12 said:
Question:
Find the volume of the solid obtained by rotating the region
bounded by the given curves about the specified line. Sketch the
region, the solid, and a typical disk or washer.
y2 = x; x = 2y; about the y-axis

Solution: http://www.slader.com/textbook/9780538497909-stewart-calculus-early-transcendentals-7th-edition/438/exercises/9/

Now, when I solved the question, I multiplies the entire integral by 2 since the limits I took were from -2 to 2.
This doesn't make any sense. The two curves intersect at (0, 0) and (4, 2). What does your integral look like?
Phys12 said:
However, in the answer, they have considered only the values above the x-axis and not below. Why is it so? It doesn't specify in the question that we need to find the area of only that region which lies above the x-axis.
The region that is being revolved around the y-axis is completely above the x-axis, except the point (0, 0). Did you sketch the region that is being revolved?
 
Mark44 said:
This doesn't make any sense. The two curves intersect at (0, 0) and (4, 2). What does your integral look like?

The region that is being revolved around the y-axis is completely above the x-axis, except the point (0, 0). Did you sketch the region that is being revolved?
Okay, I got it. Just one question, when we plot x=y^2, why don't we plot it below the x-axis as well? I know that it will cease to be a function, but we can values for x when we take y and -y, right? (That's what got me confused, I plotted y^2=x as y=x^2 but turned 90 degrees. Which made me draw another line below the x-axis as well for some reason-- stupid me!)
 
Phys12 said:
Okay, I got it. Just one question, when we plot x=y^2, why don't we plot it below the x-axis as well?
You do plot a part of it below the x-axis. But, the region that is being revolved is between this curve and the line x = 2y. That region is completely in the first quadrant.
Phys12 said:
I know that it will cease to be a function, but we can values for x when we take y and -y, right? (That's what got me confused, I plotted y^2=x as y=x^2 but turned 90 degrees. Which made me draw another line below the x-axis as well for some reason-- stupid me!)
 
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