How do you use the triple integral to find the volume of the solid bounded by the surface #z=sqrt y# and the planes x+y=1, x=0, z=0?

Answer 1

4/15 cubic units

The surface of this solid comprises three planar sides (horizontal) z =

0 and (vertical) x = 0 and x + y +1 and a part, in the first octant, of the

(parabolic cross-sectional) cylinder z = #sqrt y#. This cylinder is

symmetrical about the xy-plane.

as z = #sqrt y, y>=0#..Being bounded below by z = 0, z does not
take negative #sqrt y# values.

Now, the volume of this solid in the first octant is

#V=int int int dz dy dx#, for z from 0 to #sqrt y#, y from o to 1-x and x

from 0 to 1.

Integrating with respect to z first, in order,,

#V=int inty^(1/2)dy dx#, y from 0 to 1-x and x from 0 to 1#
#=(2/3)int(1-x)^(3/2)dx#, x from 0 to 1
#=(2/3)(-2/5)(0-1)#
#=4/15# cubic units.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#4/15#

first and key, because of the shape of #z = sqrt y# the plane y=0 is a boundary, in addition to the given x = 0, z = 0.

so we are in the first octant for all of this.

the further constraint is the plane x + y = 1

this is the best drawing i can muster

the yellow bit is the area over which we are integrating z(x,y)

but as a triple integral you would write:

#int_{x=0}^{1} \ int_{y=0}^{1-x} \ int_{z = 0}^{sqrt(y)} \ dz \ dy \ dx#

#= int_{x=0}^{1} \ int_{y=0}^{1-x} [z]_{z = 0}^{sqrt(y)} \ dy \ dx#

#= int_{x=0}^{1} \ int_{y=0}^{1-x} sqrt(y) \ dy \ dx#

#= int_{x=0}^{1} \ [2/3 y^{3/2}]_{y=0}^{1-x} \ dy \ dx#

#= int_{x=0}^{1} \ 2/3 (1-x)^{3/2} \ dx#

#= \ [- 4/15 (1-x)^{5/2} ]_{x=0}^{1}#

# = 4/15#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find the volume of the solid bounded by the surface (z = \sqrt{y}) and the planes (x+y=1), (x=0), and (z=0), we set up the triple integral over the region bounded by these surfaces.

The region of integration in this case is a triangular region in the xy-plane, bounded by the lines (x=0), (x+y=1), and the y-axis.

We can express the limits of integration as follows:

For (z), it ranges from 0 to (\sqrt{y}). For (y), it ranges from 0 to 1. For (x), it ranges from 0 to (1-y).

The integral setup is:

[ \iiint_V dz , dy , dx ]

with limits of integration:

[ 0 \leq z \leq \sqrt{y}, \quad 0 \leq y \leq 1, \quad 0 \leq x \leq 1 - y ]

Finally, we integrate over these limits to find the volume:

[ \int_0^1 \int_0^{1-y} \int_0^{\sqrt{y}} dz , dx , dy ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7