# Thread: integration, area and volume

1. ## 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

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

2. 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

I am way off, can someone please show me where I'm going wrong.
Thank you
Hmmm... The derivative of $\displaystyle \arcsin{x}$ is $\displaystyle \frac{1}{\sqrt{1-x^2}}$. So..... the integral of $\displaystyle \frac{1}{\sqrt{1-x^2}}$ is....

3. 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

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.

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

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

4. 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.

5. Originally Posted by Chris L T521
a) is correct.

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

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

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

6. 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.

7. 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: $\displaystyle V=2\pi r h$.

In our case, $\displaystyle h(x)$ (or $\displaystyle h(y)$) is the height of each shell (its usually the function you're considering rotating about a certain axis or line. In our case, its $\displaystyle \frac{1}{\sqrt{1-x^2}}$). $\displaystyle r(x)$ (or $\displaystyle r(y)$) 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, $\displaystyle r(x)=x$ for $\displaystyle x\in\left[0,1/2\right]$.

Now when we put this all together, we have $\displaystyle 2\pi\int_0^{\frac{1}{2}}r(x)h(x)\,dx=2\pi\int_0^{\ frac{1}{2}}x\cdot\frac{1}{\sqrt{1-x^2}}\,dx$, 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.

8. 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: $\displaystyle V=2\pi r h$.

In our case, $\displaystyle h(x)$ (or $\displaystyle h(y)$) is the height of each shell (its usually the function you're considering rotating about a certain axis or line. In our case, its $\displaystyle \frac{1}{\sqrt{1-x^2}}$). $\displaystyle r(x)$ (or $\displaystyle r(y)$) 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, $\displaystyle r(x)=x$ for $\displaystyle x\in\left[0,1/2\right]$.

Now when we put this all together, we have $\displaystyle 2\pi\int_0^{\frac{1}{2}}r(x)h(x)\,dx=2\pi\int_0^{\ frac{1}{2}}x\cdot\frac{1}{\sqrt{1-x^2}}\,dx$, 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!