Answer to Question #100913 in Calculus for Jeffrey Chen

Question

1. Use a double integral to find the volume V of the solid that is common to cylinders x^2 + y^2 = 81 and x^2 + z^2 = 81. Find the exact number and no tolerance

2. Use a triple integral to find the volume of the solid in the first octant bounded by the coordinate planes and the plane 9x+18y+8z = 144.

3. Use a triple integral to find the volume of the solid bounded by the surface y=x^2 and the planes x+z=8 and z=0.Give the exact answer in the form of a fraction.

Answer 1

1.$Volume = 8\int ^9_0\int^{\sqrt{81-x^2}}_0\int^{\sqrt{81-x^2}}_0dydzdx$

$Volume=8\int^9_0(\sqrt{81-x^2})^2 dx$

$Volume=8\int^9_0(81-x^2 )dx$

$Volume=8[81x-x^3/3]^9_0$

$Volume=8(81×9-81×3)$

$Volume=81×6×8=3888(Ans)$

2.

$Volume=\int ^{16}_0\int^{8-x/2}_0(16-(9/8)x-(9/4)y)dydx$

$Volume=\int ^{16}_0[(16y-(9/8)yx-(9/8)y^2)]^{8-x/2}_0$

$Volume=\int ^{16}_0[(16(8-x/2)-(9/8)(8x-x^2/2) -(9/8)(8-x/2)^2)]dx$

$Volume=[-4( 8-x/2)^2-(9/8)(4x^2-x^3/6)+(3/16)(8-x/2)^3]^{16}_0=224(ans)$

3.$Volume= \int^{64}_0\int_ {\sqrt{y}}^8\int_0^{8-x}dzdxdy$

$Volume= \int^{64}_0\int_ {\sqrt{y}}^8(8-x)dxdy$

$Volume= \int^{64}_0[(8x-x^2/2)]^8_{√y}dy$

$Volume= \int^{64}_0(32-(8√y)+y/2)dy$

$Volume= (32y-(8y^{3/2})+y^2/4)]^{64}_{0}$

$Volume= (32×64-8^4+64^2/4)]=1024(Ans)$