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 $y = 16$ ?

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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.

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

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

$V=(1024pi)/5$

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