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?
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?
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?
That is, of course, a sphere with radius 5 so you could use the formula $\displaystyle V= \frac{4}{3}\pi r^3= \frac{500}{3}\pi$ as a check.
There are many ways to integrate this. The simplest is probably to use "disks" perpendicular to the y-axis. Solving $\displaystyle x^2+ y^2= 25$ for x, $\displaystyle x= \sqrt{25- y^2}$ in the first quadrant so a disk at a specific y would have radius from x= 0 to $\displaystyle x= \sqrt{25- y^2}$. The area of such a disk is $\displaystyle pi r^2=\pi (25- y^2)$. Imagining it to have thickness $\displaystyle \Delta y$ (since the thickness is measured along the y-axis), its volume would be $\displaystyle \pi (25- y^2)\Delta y$. The volume of all of those disks is the sum $\displaystyle sum \pi (25- y^2)\Delta y$ which, in the limit as $\displaystyle \Delta y$ goes to 0, becomes the integral $\displaystyle \int_{y= -5}^5 \pi (25- y^2) dy$.
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 $\displaystyle 2y= 2\sqrt{25- x^2)$ (2 because it goes both up to $\displaystyle \sqrt{25- x^2}$ and down to $\displaystyle -\sqrt{25- x^2}$. Imagining that flattened out to a rectangle with length $\displaystyle 2\pi x$, the circumference of the cylinder, and height $\displaystyle 2\sqrt{25- x^2}$, it would have an area of $\displaystyle 4\pi x\sqrt{25- x^2}$. Taking the thickness of that cylinder to be $\displaystyle \Delta x$ (since the thickness is now measured along the y-axis), the volume of such a cylinder would be $\displaystyle 4\pi x\sqrt{25- x^2}\Delta x$ and the volume of all the cylinders together would be $\displaystyle \sum 4\pi x\sqrt{25- x^2}\Delta x$. Taking the limit as $\displaystyle \Delta x$ goes to 0, that becomes the integral $\displaystyle \int_{x=0}^5 4\pi x\sqrt{25- x^2}dx$.
Try integrating both ways and see if you don't get $\displaystyle 500\pi/3$. Do you see why the first integral is from y= -5 to 5 and the second from x= 0 to 5?
$\displaystyle x^2+y^2=25 \Leftrightarrow x^2+y^2=5^2$. It's a central circle with radius $\displaystyle r=5$. Rotating it around the y-axis would give a sphere.
Volume of the solid of revolution generated with a curve $\displaystyle y=f(x), x\in[a,b]$ revolving around the y-axis can be calculated using the following expression $\displaystyle V_y=2\pi\int\limits_{a}^{b}x\cdot f(x) dx$
Rotating only the quarter of a circle (one in the first quadrant)
$\displaystyle y=\sqrt{25-x^2}, x\in[0,5]$
would give you one half of the solid of revolution (and half of the volume needed), therefore you should multiply the result by 2.
$\displaystyle V=2\cdot 2\pi \int\limits_{0}^{5}x\cdot\sqrt{25-x^2}dx$