Volume of revolution in the first quadrant?!

  • Thread starter CAH
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  • #1
CAH
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Homework Statement


Find the volumes of the solid formed when each of the areas in the following perform one revolution about the X axis...
Question: The volume line in the first quadrant and bounded by the curve y=x^3 and the line y=3x+2.

Homework Equations


Volume of revolution about X-axis: V=pi*integral(y^2) dx. ('b' upper, and 'a' lower limit)

The Attempt at a Solution


Ok so I can find the volume (I think) of the whole system, but I can't just find the volume in the first quadrant.
V=pi*integral[(3x+2)^2 - (x^3)^2] between 2 and -1. So this is area under line minus area under curve, between there intersections. I don't know how to find the area in the first quadrant alone though!
I got V= 264pi/7 (the whole system) and the answer in my textbook is 56pi/5. Also sometimes the textbooks are wrong.
Thanks
 

Answers and Replies

  • #2
34,675
6,386

Homework Statement


Find the volumes of the solid formed when each of the areas in the following perform one revolution about the X axis...
Question: The volume line in the first quadrant and bounded by the curve y=x^3 and the line y=3x+2.

Homework Equations


Volume of revolution about X-axis: V=pi*integral(y^2) dx. ('b' upper, and 'a' lower limit)

The Attempt at a Solution


Ok so I can find the volume (I think) of the whole system, but I can't just find the volume in the first quadrant.
V=pi*integral[(3x+2)^2 - (x^3)^2] between 2 and -1.
In the first quadrant, x > 0, so your integral should not be from - 1 to 2. The region that is being revolved is bounded above by the line y = 3x + 2, below by the curve y = x3, and on the left by the y-axis (because you're interested only in what's happening in Quadrant 1).
CAH said:
So this is area under line minus area under curve, between there intersections. I don't know how to find the area in the first quadrant alone though!
I got V= 264pi/7 (the whole system) and the answer in my textbook is 56pi/5. Also sometimes the textbooks are wrong.
Thanks
 
  • #3
CAH
48
0
Yea i did the 0-2 boundary, must have accidentally wrote -1! But I still have the area beneath the X axis in my equation and I don't know how to take it away so it's only the area in the first quad left...?
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
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767

Homework Statement


Find the volumes of the solid formed when each of the areas in the following perform one revolution about the X axis...
Question: The volume line in the first quadrant and bounded by the curve y=x^3 and the line y=3x+2.

Homework Equations


Volume of revolution about X-axis: V=pi*integral(y^2) dx. ('b' upper, and 'a' lower limit)
Remember, that formula is for the area between the ##x## axis and the function, rotated. You need the formula$$
A =\pi \int_a^b y_{upper}^2 - y_{lower}^2~dx$$to get the volume of the area between the curves rotated.
 

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