Revolution Volume

[Wakigawa]

Hello! Sorry if there is any typos, english is not my native language.

Being "a" a real and positive number. The volume of the solid obtained by spinning around the x axis the region of the first quadrant limited by y=ax, the x axis and the a²x²+y²=2a² ellipsis is equal to 4π{sqrt{2}-1}. What is the value of a?​

First i tried to visualize the graph, by plotting it. But now i am clueless as to how to find the values of a. Should i equal y=ax to the elipsis to find the intersection point?
I tried isolating Y and working with the positive part of the square root, and found sqrt{3}. Is this correct?
Thank you!

 

[skeeter]

intersection in quad 1 …

[imath]x= \sqrt{2-x^2} \implies x = 1[/imath]

[math]V = \pi a^2 \bigg[ \int_0^1 x^2 \, dx + \int_1^{\sqrt{2}} (2-x^2) \, dx \bigg][/math]

 

[Dr.Peterson]

Hello! Sorry if there is any typos, english is not my native language.

Being "a" a real and positive number. The volume of the solid obtained by spinning around the x axis the region of the first quadrant limited by y=ax, the x axis and the a²x²+y²=2a² ellipsis is equal to 4π{sqrt{2}-1}. What is the value of a?​

First i tried to visualize the graph, by plotting it. But now i am clueless as to how to find the values of a. Should i equal y=ax to the elipsis to find the intersection point?
I tried isolating Y and working with the positive part of the square root, and found sqrt{3}. Is this correct?
Thank you!
View attachment 28272

If you're looking for the intersection of [imath]y = ax[/imath] and [imath]a^2x^2 + y^2 = 2a^2[/imath], just replace y with ax and solve for x. The answer is very simple. (What are you saying is equal to [imath]\sqrt{3}[/imath]?)

Then you have to find the volume.

 

[Eugenio]

I have plenty of time on this very cold morning here in Argentina, so I'm going to solve this problem step by step for posterity.

To find the intersection between [imath]y=ax[/imath] and [imath]a^2x^2+y^2=2a^2[/imath] we do

[math]a^2x^2+(ax)^2=2a^2[/math]
[math]2a^2x^2=2a^2[/math]
[math]x^2=1 \Rightarrow x=1[/math]
We choose the positive solution because we are in the first quadrant

The intersection between y = ax and the x axis is x = 0, and for the ellipsis we replace y = 0 on its equation

[math]a^2x^2=2a^2 \Rightarrow x=\sqrt{2}[/math]
The volume of a solid of revolution, obtained by spinning the graph of f(x) around the x axis in an interval [a; b], can be calculated with

[math]V = \pi \int_{a}^{b}\left [ f(x) \right ]^2dx[/math]
for [imath]x \epsilon [0; 1] \rightarrow [f(x)]^2=a^2x^2[/imath] and for [imath]x \epsilon [1; \sqrt{2}] \rightarrow [f(x)]^2=2a^2-a^2x^2=a^2(2-x^2)[/imath] so,

[math]V = \pi \int_{0}^{1} a^2x^2 dx + \pi \int_{1}^{\sqrt{2}} a^2(2-x^2) dx = \pi a^2\left (\int_{0}^{1} x^2 dx + \int_{1}^{\sqrt{2}} (2-x^2) dx \right ) =\pi a^2\frac{4}{3}(\sqrt{2}-1)[/math]
Then, [imath]a=\sqrt{3}[/imath]