# Math Help - application of definite integral

1. ## application of definite integral

"A round hole of radius $\sqrt{3}$ is bored through the center of a sphere of radius 2 feet. Find the volume of the piece cut out."

I have a vague idea on how to do this. $\int_{-\sqrt{3}}^{\sqrt{3}} 2\Pi r^{2}+2\Pi rh$. Integral of the surface area. Except, I don't know what to do with the extra variable.

2. Originally Posted by TheMathTham
"A round hole of radius $\sqrt{3}$ is bored through the center of a sphere of radius 2 feet. Find the volume of the piece cut out."

I have a vague idea on how to do this. $\int_{-\sqrt{3}}^{\sqrt{3}} 2\Pi r^{2}+2\Pi rh$. Integral of the surface area. Except, I don't know what to do with the extra variable.
If you think about the diagram, you'll notice that the h is simply the radius of the sphere, 2 feet. Now, you only have one variable.

3. Originally Posted by mathemagister
If you think about the diagram, you'll notice that the h is simply the radius of the sphere, 2 feet. Now, you only have one variable.
I did that, though I didn't get the right answer, I did notice something else. The cylinder would not have a flat top, it would have a bulbous top on each side of the cylinder because of the sphere. That would mean that the cylinder's height would not be 2, it would be (2-2r). The two bulbous lids would have a height of r. But when I tried to add those in, I get a negative number? The correct answer is $\frac{28\Pi }{3}\approx 29.3215$. My logic is starting to confuse me...

4. Originally Posted by TheMathTham
"A round hole of radius $\sqrt{3}$ is bored through the center of a sphere of radius 2 feet. Find the volume of the piece cut out."

I have a vague idea on how to do this. $\int_{-\sqrt{3}}^{\sqrt{3}} 2\Pi r^{2}+2\Pi rh$. Integral of the surface area. Except, I don't know what to do with the extra variable.
If you rotate the region enclosed by the circle $x^2+y^2=4$ and the y-axis and the line $x=\sqrt{3}$ about the y-axis, the resulting solid would what is bored out of the sphere.

Using shells, the volume would be

$\int_0^{\sqrt{3}}(2\pi{x})(2\sqrt{4-x^2})\,dx$

5. Originally Posted by TheMathTham
I did that, though I didn't get the right answer, I did notice something else. The cylinder would not have a flat top, it would have a bulbous top on each side of the cylinder because of the sphere. That would mean that the cylinder's height would not be 2, it would be (2-2r). The two bulbous lids would have a height of r. But when I tried to add those in, I get a negative number? The correct answer is $\frac{28\Pi }{3}\approx 29.3215$. My logic is starting to confuse me...
Okay, I did some researching on the almighty Internet and found this...

Draw a 2-dimensional picture of a cross-section through the middle. Look at the right hand side. You have a 2-dimensional object bounded on the right by the curve $x^2 + y^2 = 4$ and on the left by the line $x = \sqrt {3}$. The points where these curves intersect are $(\sqrt {3},1)$ and $(\sqrt {3}, - 1)$. We want the volume of the solid of revolution formed by rotating this about the y axis.

Method 1 (the shell method):

Slice the 2-dimensional object into vertical strips. Each such vertical strip is located at position x, has width dx, and height $2\sqrt {4 - x^2}$. Rotating this strip about the vertical axis gives a thin cylindrical shell. Slice this open and lay it out flat. The result is a rectangular slap of width $2\pi x$ (the circumference of the shell), height $2\sqrt {4 - x^2}$, and thickness dx so $dV = 4\pi x\sqrt {4 - x^2}\,dx$ and

$V = \int_{\sqrt {3}}^24\pi x\sqrt {4 - x^2}\,dx$

Method 2 (the "washer" method):

Slice the 2-dimensional object into horizontal strips. Each such horizontal strip is located at position y and has width dy. Its left endpoint is at $x = \sqrt {3}$ and its right endpoint is at $x = \sqrt {4 - y^2}$. Rotating this strip about the vertical axis gives a thin flat "washer" of thickness dy. The inner radius of this (annulus) is $\sqrt {3}$ and the outer radius is $\sqrt {4 - y^2}$. The volume is

$dV = \pi\left(\left(\sqrt {4 - y^2}\right)^2 - \left(\sqrt {3}\right)^2\right)dy = \pi(1 - y^2)dy$

$V = \int_{ - 1}^1 \pi(1 - y^2)dy$

I'm sure you know how to solve either of these integrals.

6. Oh okay. That makes sense. Ione's equation is correct. I was unfamiliar with the equations that produce semi-circles. I'm confused on why the height is $2\sqrt{4-x^{2}}$ though. I would have thought that it would be $f(x) - g(x)$, aka $\sqrt{4-x^{2}} - \sqrt{3}$

7. Originally Posted by TheMathTham
Oh okay. That makes sense. Ione's equation is correct. I was unfamiliar with the equations that produce semi-circles. I'm confused on why the height is $2\sqrt{4-x^{2}}$ though. I would have thought that it would be $f(x) - g(x)$, aka $\sqrt{4-x^{2}} - \sqrt{3}$
Your slices go from the top part of the circle to the bottom part of the circle so the length of each slice is twice the distance from the x-axis to the top part of the circle.

8. Originally Posted by mathemagister
Okay, I did some researching on the almighty Internet and found this...

Draw a 2-dimensional picture of a cross-section through the middle. Look at the right hand side. You have a 2-dimensional object bounded on the right by the curve $x^2 + y^2 = 4$ and on the left by the line $x = \sqrt {3}$. The points where these curves intersect are $(\sqrt {3},1)$ and $(\sqrt {3}, - 1)$. We want the volume of the solid of revolution formed by rotating this about the y axis.

Method 1 (the shell method):

Slice the 2-dimensional object into vertical strips. Each such vertical strip is located at position x, has width dx, and height $2\sqrt {4 - x^2}$. Rotating this strip about the vertical axis gives a thin cylindrical shell. Slice this open and lay it out flat. The result is a rectangular slap of width $2\pi x$ (the circumference of the shell), height $2\sqrt {4 - x^2}$, and thickness dx so $dV = 4\pi x\sqrt {4 - x^2}\,dx$ and

$V = \int_{\sqrt {3}}^24\pi x\sqrt {4 - x^2}\,dx$

Method 2 (the "washer" method):

Slice the 2-dimensional object into horizontal strips. Each such horizontal strip is located at position y and has width dy. Its left endpoint is at $x = \sqrt {3}$ and its right endpoint is at $x = \sqrt {4 - y^2}$. Rotating this strip about the vertical axis gives a thin flat "washer" of thickness dy. The inner radius of this (annulus) is $\sqrt {3}$ and the outer radius is $\sqrt {4 - y^2}$. The volume is

$dV = \pi\left(\left(\sqrt {4 - y^2}\right)^2 - \left(\sqrt {3}\right)^2\right)dy = \pi(1 - y^2)dy$

$V = \int_{ - 1}^1 \pi(1 - y^2)dy$

I'm sure you know how to solve either of these integrals.
Do you mean that mine were wrong? Do you know how it that might be?
Thanks