How do you find the volume of the region bounded above by the line #y = 16#, below by the curve #y = 16-x^2#, and on the right by the line #x = 4# about the line #y = 16#?

Question

8408 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-21T13:04:46

#(1024pi)/5#

Intersection of the curve #y=16-x^2# and x-axis is:
#0=16-x^2 <=> x^2=16 <=> x=4 vv x=-4#
It's the same bounding line as the one given in the task.

#V=piint_0^4(r_1^2-r_2^2)dx#

where #r_1# and #r_2# are outer and inner radius of the solid, respectively.

#r_1=16-(16-x^2)=16-16+x^2=x^2#
#r_2=16-16=0#

#V=piint_0^4((x^2)^2-0^2)dx=piint_0^4 x^4dx=pi x^5/5|_0^4#

#V=(1024pi)/5#