Thread: Calc volume / area between curve help:

1. Calc volume / area between curve help:

Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.

For #1, I got 16.5.

2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

For some reason I keep getting a negative answer here. That can't be right for a volume problem. Also, another random note: it does say rotate around y-axis, but even if you did it around x-axis, do you get the same answer?

2. Originally Posted by XFactah416 Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.

For #1, I got 16.5.
How did you get this answer? Originally Posted by XFactah416 2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

$\displaystyle V = \pi\int_1^4 y^2~dx$

3. Originally Posted by pickslides How did you get this answer?

$\displaystyle V = \pi\int_1^4 y^2~dx$
For #1:

I chose to use horizontal strips, so I converted y = x + 5 into x = y-5.

Thus, I had an interval of -1 to 2, and it was [right curve - left curve] dy. My right curve was y^2, and my left curve was (y-5).

After all the math, I got 16.5. I'm guessing it's incorrect?

For #2, are you just saying to square (2+5x^2) ^ (-1/2) and then take the integral of that? I have to rotate around the y-axis in this problem. I was under the impression that when you rotate around the y axis, you must convert the equation into an x = y equation.

4. Originally Posted by XFactah416 For #1:

I chose to use horizontal strips, so I converted y = x + 5 into x = y-5.

Thus, I had an interval of -1 to 2, and it was [right curve - left curve] dy. My right curve was y^2, and my left curve was (y-5).

After all the math, I got 16.5. I'm guessing it's incorrect?
O.k, it really depends on the width of your horizontal strips, could be correct. Originally Posted by XFactah416 For #2, are you just saying to square (2+5x^2) ^ (-1/2) and then take the integral of that? I have to rotate around the y-axis in this problem. I was under the impression that when you rotate around the y axis, you must convert the equation into an x = y equation.
yep, square it, integrate substituting in the terminals and multiply the result by pi.

5. OK.

Thank you! However, I'm guessing my theory is incorrect about rotation around y-axis means that we have to have an equation of x = y?

6. Originally Posted by XFactah416 Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.

For #1, I got 16.5. <<<<<<<< OK 2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

For some reason I keep getting a negative answer here. That can't be right for a volume problem. Also, another random note: it does say rotate around y-axis, but even if you did it around x-axis, do you get the same answer?
1. The general form of the volume which you get by rotation about the y-axis is:

$\displaystyle V_{roty} = \int_{f(a)}^{f(b)} \pi x^2 dy$

2. To calculate the volume it isn't necessary to determine x but :

$\displaystyle y = \dfrac1{\sqrt{2+5x^2}}~\implies~x^2 = \dfrac1{5y^2} - \dfrac25$

If y = f(x) you'll get $\displaystyle f(1) = \frac1{\sqrt{7}}$ and $\displaystyle f(4) = \frac1{\sqrt{22}}$

The complete equation to calculate the volume is

$\displaystyle V_{roty}=\int_{\tfrac1{\sqrt{22}}}^{\tfrac1{\sqrt{ 7}}} \left( \pi \cdot \left(\dfrac1{5y^2} - \dfrac25 \right) \right) dy$

3. I've got an approximate value of the volume of 1.07765

4. The volume of rotation at rotation about the x-axis is calculated by:

$\displaystyle V_{rotx} = \pi \int_a^b y^2 dx$

In genaral the volumes $\displaystyle V_{rotx}$ and $\displaystyle V_{roty}$ are unequal.

7. Originally Posted by XFactah416 Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.
What region are you talking about? The line y= x+ 5 does NOT intersect the curve $\displaystyle x= y^2$ between y= -1 and y= 2.

For #1, I got 16.5.

2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

For some reason I keep getting a negative answer here. That can't be right for a volume problem. Also, another random note: it does say rotate around y-axis, but even if you did it around x-axis, do you get the same answer?[/QUOTE]

8. Originally Posted by HallsofIvy What region are you talking about? The line y= x+ 5 does NOT intersect the curve $\displaystyle x= y^2$ between y= -1 and y= 2.

....
When I checked the result I assumed that the region was meant which is coloured in grey. And then the posted result is OK.

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