# integration, area and volume

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• Jul 11th 2009, 06:34 PM
slaypullingcat
integration, area and volume
a) Find the area enclosed between y=1/sqrt(1-x^2), the x axis and the lines x=0 and x=1/2.

integrating y I got [sin^-1(x)] between limits 1/2 and 0 which equals
=[sin^-1(1/2)]-[sin^-1(0)]
=pi/6 units^2

b) This area is rotated about the y-axis. Find the exact volume of the soild formed.

This is what I am thinking...........use V= pi integral x^2 dy with limits pi/6 and 0. I rearranged y from part a) to make x^2 the subject and got x^2=1-y^-2.

Anywho I integrated and got V=5pi/4 units^3

The answer is pi/6(7sqrt(3)-12)units^3

I am way off, can someone please show me where I'm going wrong.
Thank you
• Jul 11th 2009, 06:38 PM
VonNemo19
Quote:

Originally Posted by slaypullingcat
a) Find the area enclosed between y=1/sqrt(1-x^2), the x axis and the lines x=0 and x=1/2.

integrating y I got [sin^-1(x)] between limits 1/2 and 0 which equals
=[sin^-1(1/2)]-[sin^-1(0)]
=pi/6 units^2

b) This area is rotated about the y-axis. Find the exact volume of the soild formed.

This is what I am thinking...........use V= pi integral x^2 dy with limits pi/6 and 0. I rearranged y from part a) to make x^2 the subject and got x^2=1-y^-2.

Anywho I integrated and got V=5pi/4 units^3

The answer is pi/6(7sqrt(3)-12)units^3

I am way off, can someone please show me where I'm going wrong.
Thank you

Hmmm... The derivative of $\arcsin{x}$ is $\frac{1}{\sqrt{1-x^2}}$. So..... the integral of $\frac{1}{\sqrt{1-x^2}}$ is....(Wondering)
• Jul 11th 2009, 06:53 PM
Chris L T521
Quote:

Originally Posted by slaypullingcat
a) Find the area enclosed between y=1/sqrt(1-x^2), the x axis and the lines x=0 and x=1/2.

integrating y I got [sin^-1(x)] between limits 1/2 and 0 which equals
=[sin^-1(1/2)]-[sin^-1(0)]
=pi/6 units^2

b) This area is rotated about the y-axis. Find the exact volume of the soild formed.

This is what I am thinking...........use V= pi integral x^2 dy with limits pi/6 and 0. I rearranged y from part a) to make x^2 the subject and got x^2=1-y^-2.

Anywho I integrated and got V=5pi/4 units^3

The answer is pi/6(7sqrt(3)-12)units^3

I am way off, can someone please show me where I'm going wrong.
Thank you

a) is correct.

Now for b), I would suggest the method of cylindrical shells.

$V=2\pi\int_0^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}\,dx$.

However, I get a different solution that the one you claim it to be... (Wondering)
• Jul 11th 2009, 06:59 PM
VonNemo19
Quote:

Originally Posted by slaypullingcat
[sin^-1(x)]

Latex, latex, latex, my friend. I had trouble with your association here. If you wrap math tags ( the sigma button ) around your expression there, you'd see the trouble.
• Jul 11th 2009, 07:01 PM
slaypullingcat
Quote:

Originally Posted by Chris L T521
a) is correct.

Now for b), I would suggest the method of cylindrical shells.

$V=2\pi\int_0^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}\,dx$.

However, I get a different solution that the one you claim it to be... (Wondering)

that answer is coming from the textbook, I just double checked it. Thanks for your post, its a bit over my head but I appreciate your time.
• Jul 11th 2009, 07:05 PM
slaypullingcat
Quote:

Originally Posted by VonNemo19
Latex, latex, latex, my friend. I had trouble with your association here. If you wrap math tags ( the sigma button ) around your expression there, you'd see the trouble.

Sorry, I know I should make the effrot. I'm two weeks out from my final exam and I'm putting all my time into study.
• Jul 11th 2009, 07:18 PM
Chris L T521
Quote:

Originally Posted by slaypullingcat
that answer is coming from the textbook, I just double checked it. Thanks for your post, its a bit over my head but I appreciate your time.

Its hard to get used to all the different techniques.

Basically, it comes from the definition of the volume of a cylinder: $V=2\pi r h$.

In our case, $h(x)$ (or $h(y)$) is the height of each shell (its usually the function you're considering rotating about a certain axis or line. In our case, its $\frac{1}{\sqrt{1-x^2}}$). $r(x)$ (or $r(y)$) is the radius "distance" from the axis of rotation.

Since we want to revolve around the y axis, we measure the radius of each shell from 0 to 1/2 (in otherwords, $r(x)=x$ for $x\in\left[0,1/2\right]$.

Now when we put this all together, we have $2\pi\int_0^{\frac{1}{2}}r(x)h(x)\,dx=2\pi\int_0^{\ frac{1}{2}}x\cdot\frac{1}{\sqrt{1-x^2}}\,dx$, which is the integral I gave you.

I hope this clarifies how I came up with that integral.

See this pdf for more of an explanation on cylindrical shells.
• Jul 11th 2009, 07:58 PM
slaypullingcat
Quote:

Originally Posted by Chris L T521
Its hard to get used to all the different techniques.

Basically, it comes from the definition of the volume of a cylinder: $V=2\pi r h$.

In our case, $h(x)$ (or $h(y)$) is the height of each shell (its usually the function you're considering rotating about a certain axis or line. In our case, its $\frac{1}{\sqrt{1-x^2}}$). $r(x)$ (or $r(y)$) is the radius "distance" from the axis of rotation.

Since we want to revolve around the y axis, we measure the radius of each shell from 0 to 1/2 (in otherwords, $r(x)=x$ for $x\in\left[0,1/2\right]$.

Now when we put this all together, we have $2\pi\int_0^{\frac{1}{2}}r(x)h(x)\,dx=2\pi\int_0^{\ frac{1}{2}}x\cdot\frac{1}{\sqrt{1-x^2}}\,dx$, which is the integral I gave you.

I hope this clarifies how I came up with that integral.

See this pdf for more of an explanation on cylindrical shells.

Thank you very much. I get it now. Thank you!