Geometric Applications of Definite Integrals.

**Ahplym** · December 10th 2010, 11:49 AM

Volume of solid of revolution the rotation curve:
x^2 + y^2 = 25 around the y-axis

Outcome=500*PI/3

I have to be calculated using definite integral. But I don't really know how to calculate it. Can someone help me please?

**BariMutation** · December 10th 2010, 12:08 PM

You'll need a triple integral, and in my opinion, it'd be easiest to do this using spherical coordinates. If any of this is confusing, let me know and I can simplify it using only rectangular coordinates.

You'll have a triple integral in the order of d(rho)d(phi)d(theta). The limits on theta will be 0 to 2pi (because it's a sphere), the limits on phi will be 0 to pi/2 (the angle it forms with the z axis), and the limits on rho will be 0 to 5 (the distance from the origin to the curve). In the integrand will be 1, since you're finding volume. Understand, or do I need to use rectangular coordinates?

**HallsofIvy** · December 10th 2010, 12:09 PM

That is, of course, a sphere with radius 5 so you could use the formula $V= \frac{4}{3}\pi r^3= \frac{500}{3}\pi$ as a check.

There are many ways to integrate this. The simplest is probably to use "disks" perpendicular to the y-axis. Solving $x^2+ y^2= 25$ for x, $x= \sqrt{25- y^2}$ in the first quadrant so a disk at a specific y would have radius from x= 0 to $x= \sqrt{25- y^2}$ . The area of such a disk is $pi r^2=\pi (25- y^2)$ . Imagining it to have thickness $\Delta y$ (since the thickness is measured along the y-axis), its volume would be $\pi (25- y^2)\Delta y$ . The volume of all of those disks is the sum $sum \pi (25- y^2)\Delta y$ which, in the limit as $\Delta y$ goes to 0, becomes the integral $\int_{y= -5}^5 \pi (25- y^2) dy$ .

Another, slightly harder method would be to use "shells". Imagine the sphere made up of many small cylinders with axis along tye y-axis and radius "x". Each such cylinder would have radius x and height $2y= 2\sqrt{25- x^2)$ (2 because it goes both up to $\sqrt{25- x^2}$ and down to $-\sqrt{25- x^2}$ . Imagining that flattened out to a rectangle with length $2\pi x$ , the circumference of the cylinder, and height $2\sqrt{25- x^2}$ , it would have an area of $4\pi x\sqrt{25- x^2}$ . Taking the thickness of that cylinder to be $\Delta x$ (since the thickness is now measured along the y-axis), the volume of such a cylinder would be $4\pi x\sqrt{25- x^2}\Delta x$ and the volume of all the cylinders together would be $\sum 4\pi x\sqrt{25- x^2}\Delta x$ . Taking the limit as $\Delta x$ goes to 0, that becomes the integral $\int_{x=0}^5 4\pi x\sqrt{25- x^2}dx$ .

Try integrating both ways and see if you don't get $500\pi/3$ . Do you see why the first integral is from y= -5 to 5 and the second from x= 0 to 5?

**Also sprach Zarathustra** · December 10th 2010, 12:09 PM

You can use the formula of a ball: V=(4/3) pi*R^3 (Your R is 5) :-)

$V=2\pi\int^5_0 x^2dy=2\pi\int^5_0 (25-y^2)dy=...$

**Ahplym** · December 10th 2010, 12:10 PM

It has to be calculated in rectangular coordinates.

**MathoMan** · December 10th 2010, 12:22 PM

$x^2+y^2=25 \Leftrightarrow x^2+y^2=5^2$ . It's a central circle with radius $r=5$ . Rotating it around the y-axis would give a sphere.

Volume of the solid of revolution generated with a curve $y=f(x), x\in[a,b]$ revolving around the y-axis can be calculated using the following expression $V_y=2\pi\int\limits_{a}^{b}x\cdot f(x) dx$

Rotating only the quarter of a circle (one in the first quadrant)
$y=\sqrt{25-x^2}, x\in[0,5]$
would give you one half of the solid of revolution (and half of the volume needed), therefore you should multiply the result by 2.
$V=2\cdot 2\pi \int\limits_{0}^{5}x\cdot\sqrt{25-x^2}dx$

**Ahplym** · December 10th 2010, 12:35 PM

Thank you so much all for your help.

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