Use a triple integral to compute the exact volume of the solid enclosed by 𝑦=𝑥^2, 𝑥=𝑦^2, 𝑧=𝑥+𝑦+27, and 𝑧=0.

Question

Use a triple integral to compute the exact volume of the solid enclosed by y=x², x=y², z=x+y+27, and z=0.

Answer 1

There are 2 key steps here: rewriting x as a function of y to y as a function of x and finding where f(x) and g(x) meet

Let f(x) = x²

If x =y², for positive x, then √x = y, so

g(x) = √x

Then, f(x) = x² and g(x) = √x

√x - x² = 0

√x(1 - x^3/2 ) = 0

√x = 0 or x^3/2 = 1

x = 0, 1

z ranges from 0 to x+y+27

Then

I[0, 1] I[x², √x] I[0,x + y + 27] dz dy dx =

I[0, 1] I[x², √x] z E[0,x + y + 27] dy dx =

I[0, 1] I[x², √x] x + y + 27 dy dx =

I[0, 1] xy + y²/2 + 27y E[x², √x] dx =

I[0, 1] x^3/2 + x/2 + 27 x^1/2 - (x³ + x⁴/2 + 27x²) dx =

2/5 x^5/2 + x²/4 + 18 x^3/2 - x⁴/4 - x⁵/10 - 9x³ E[0, 1] =

2/5 + 1/4 + 18 - 1/4 - 1/10 - 9 =

9 3/10