anonymous 4 years ago find the region of the solid formed by rotating the area bounded by y=3, y=sqrt(x), and the x-axis. Rotate around x=10.

1. ash2326

we have the curve y=sqrt (x)|dw:1327469538153:dw|

2. ash2326

if we have to calculate area bounded, it'll be between the curves x=y^2 and x=10 take a rectangle of length 10-y^2 and thickness dy, y will vary from y=3 to y= root 10 $\int\limits_{3}^{\sqrt{10}}( 10-y^2) dy$ now the area is rotated about x=10 so the volume of a disc= $$\pi*(10-y^2)*dy$$ radius is 10-y^2 thickness is dy y will vary from y=3 to y= $$\sqrt 10$$ Volume =$$\int\limits_{3}^{\sqrt{10}} \pi* (10-y^2)^2dy$$ solving the integral, you'll get volume =0.1712 http://www.wolframalpha.com/input/?i=integrate+%28pi*%2810-y%5E2%29%5E2%29dy+from+y%3D3+to+y%3D+10%5E%281%2F2%29