Calculus: Early Transcendentals, Chapter 6, 6.2, Section 6.2, Problem 8

y=x^2/4,y=5-x^2
Refer the attached image. Graph of y=x^2/4 is plotted in red color and graph of y=5-x^2 is plotted in blue color.
From the graph, the curves intersects at x=-2 , x=2.
Using washer method,
Inner radius of washer=x^2/4
Outer radius of washer=5-x^2
Cross sectional area of washer A(x)=pi((5-x^2)^2-(x^2/4)^2)
A(x)=pi(25+x^4-10x^2-x^4/16)
A(x)=pi(15/16x^4-10x^2+25)
Volume of the solid obtained by rotating the region bounded by the curves about the x=axis ,V=int_(-2)^2A(x)dx
V=int_(-2)^2pi(15/16x^4-10x^2+25)dx
V=pi[15/16*x^5/5-10*x^3/3+25x]_(-2)^2
V=pi[3/16*x^5-10/3*x^3+25x]_(-2)^2
V=pi((3/16*2^5-10/3*2^3+25*2)-(3/16*(-2)^5-10/3*(-2)^3+25*(-2)))
V=pi((6-80/3+50)-(-6+80/3-50))
V=pi(6-80/3+50+6-80/3+50)
V=pi(112-160/3)
V=pi(336-160)/3
V=pi(176/3)
V=(176pi)/3