Question

Use a triple integral to find the volume of the given solid:
The solid bounded by the cylinder y = x^2 and the planes z = 0 , z = 4 and y =9

Please explain step by step.

Answer 1

this problem can actually be done using only 1 integral but i will show you the longer triple integral

first determine limits on variables:
z -> 0 to 4
y -> 0 to 9
x -> -sqrty to sqrty

y = x^2 --> x = +-sqrt(y)
\[V = \int\limits_{0}^{4}\int\limits_{0}^{9}\int\limits_{-\sqrt{y}}^{\sqrt{y}} dx dy dz\]

Answer 2

would you it be:
\[V \int\limits_{0}^{4} \int\limits_{0}^{9} \int\limits_{-\sqrt{9}}^{\sqrt{9}} x^2 dx dydz \]

Answer 3

integrating to get x^3/3 evaluated at (-sqrt 9 to sqrt 9)

Answer 4

i don't think so...i guess i could be wrong setting up the triple integral
but i know the answer is 144
area inside parabola y=x^2 from -3 to 3 is 36
36*4 = 144

Answer 5

It is 144 how would i evaluate the integral if not that way?

Answer 6

\[4\int\limits_{-3}^{3}(9-x^{2}) dx\]

Answer 7

where does the 4 come from?
and that would give u = 9-x^2
du = -x^(3)/3dx
ad im unsure where to go from there

Answer 8

height of cylinder is 4, z goes from 0 to 4
no substitution needed, just use power rule (its a polynomial)

Answer 9

oh its just 9x- x^(3)/3? evaluated from 2 to -2 and then moving onto dydz with 4 in the front?