Convert to cylindrical coordinates:
[tex]x=x[/tex]
[tex]y=r\cos\theta[/tex]
[tex]z=r\sin\theta[/tex]
Then [tex]E[/tex] is the set of points [tex](r,\theta,x)[/tex] such that [tex]0\le r\le1[/tex], [tex]0\le\theta\le2\pi[/tex], and [tex]7y^2+7z^2\le x\le7[/tex] or [tex]7r^2\le x\le7[/tex].
Now
[tex]\displaystyle\iiint_E5x\,\mathrm dV=5\int_0^{2\pi}\int_0^1\int_{7r^2}^7xr\,\mathrm dx\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{245\pi}3}[/tex]
Following are the triple integral calculation to the given question:
Given:
[tex]\int \int \int_{E} 5x dv\\\\x=7y^2+7z^2\\\\x=7[/tex]
To find:
Evaluate the triple integral=?
Solution:
by using parameterization convert the cylindrical coordinates:
[tex]x=x\\\\y=r \cos \theta\\\\z= r \sin \theta\\\\[/tex]
In the given question the E is the set of points [tex](r, \theta, x)[/tex] like [tex],0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi[/tex] and [tex]7y^2+7z^2 \leq x \leq 7\ or\ 7x^2 \leq x \leq 7.[/tex]
[tex]\int \int \int_{E} 5x dv\\\\\int^{2\pi}_{0} \int^{1}_{0} \int^{7}_{7r^2} xr \ dx\ dr \ d \theta\\\\[/tex]
by solving the value we get:
[tex]=\frac{245 \pi}{3}[/tex]
Therefore, the final answer is "[tex]\bold{\frac{245 \pi}{3}}[/tex]"
Learn more:
brainly.com/question/17009873
