Answer to Question #199256 in Calculus for Vincent earl Carmo

Question #199256

Find the centroid of the solid generated if the region bounded by y = 2- x^2, x=0, and y=0 is revolved about the y-axis.


1
Expert's answer
2021-05-30T23:20:46-0400

Solution:

Given, f(x)=y=2x2f(x)=y=2-x^2




Since, it is revolving around y-axis, centroid will lie on y-axis.

So, it is of the form C(0,y,0)C(0,y,0).

y=0bf2(x)dx2abf(x)dx=02(2x2)2dx202(2x2)dxy=\dfrac{\int_0^bf^2(x)dx}{2\int_a^bf(x)dx} \\=\dfrac{\int_0^{\sqrt{2}}(2-x^2)^2dx}{2\int_0^{\sqrt{2}}(2-x^2)dx}

=02(4+x44x2)dx202(2x2)dx=\dfrac{\int_0^{\sqrt{2}}(4+x^4-4x^2)dx}{2\int_0^{\sqrt{2}}(2-x^2)dx}

=(4x+x5/54x3/3)022(2xx3/3)02=\dfrac{(4x+x^5/5-4x^3/3)_0^{\sqrt{2}}}{2(2x-x^3/3)_0^{\sqrt{2}}}

=(42+(2)5/54(2)3/3)2(22(2)3/3)=\dfrac{(4{\sqrt{2}}+({\sqrt{2}})^5/5-4({\sqrt{2}})^3/3)}{2(2{\sqrt{2}}-({\sqrt{2}})^3/3)}

=(42+(42)/58(2)/3)2(22(22)/3)=\dfrac{(4{\sqrt{2}}+(4{\sqrt{2}})/5-8({\sqrt{2}})/3)}{2(2{\sqrt{2}}-(2{\sqrt{2}})/3)}

=45=\frac45

Now, C(0,45,0)C(0,\frac45,0)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment