# Calculate the volume of a solid whose base is the ellipse # 4x^2 + y^2 = 4 # and has vertical cross sections that are square?

I got

So, it is similar to this:

An ellipse is defined as:

#x^2/a^2 + y^2/b^2 = 1#

Rewrite your ellipse equation as:

#1/4 xx (4x^2 + 1y^2 = 4)#

#=> ul(x^2/1^2 + y^2/2^2 = 1)#

Thus,

Now what I would do is:

- Obtain an equation for one-fourth of this ellipse.
- Project it along the
#x# axis from#0# to#1# using#y xx 2y# rectangles (the height spanning#0 harr (z = 2y)# ). - Multiply the resultant volume by
#4# to get the result by symmetry.(Projecting in

#0->1# is half the volume, and I only considered#(+x,+y)# values.)Try solving for

#y# in terms of#x# :#y_(+) = 2sqrt(1 - x^2)# In this case, our

#dV# , the differential volume, shall be defined as:#dV = (y xx 2y)dx = 2y^2dx = 2(2sqrt(1 - x^2))^2dx# #= 8(1 - x^2)dx# So, the projection gives

#1/4# the total volume, and the total volume is:#color(blue)(4 xx V/4) = 4 xx int_(0)^(1) 8(1 - x^2)dx# #= 32|[x - x^3/3]|_(0)^(1)# #= 32[(1 - 1^3/3) - (0 - 0^3/3)]# #= 32[1 - 1/3]# #= color(blue)(64/3 "u"^3)#

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Consider a vertical view of the base of the object.

The grey shaded area represents a top view of the square cross section. In order to find the volume of the solid we seek the volume of a generic cross sectional square "slice" and integrate over the entire base (the ellipse)

The equation of the ellipse is:

# 4x^2 + y^2 = 4 #

So for some arbitrary

# y^2=4-4x^2 #

# :. y = +-sqrt(4-4x^2) #

So for that arbitrary

# y_1 = +sqrt(4-4x^2) #

# y_2 = -sqrt(4-4x^2) #

Thus, the length of a side of an arbitrary cross sectional square slice is:

# l = y_1 - y_2 #

# \ \ = sqrt(4-4x^2) - (-sqrt(4-4x^2) ) #

# \ \ = 2sqrt(4-4x^2) #

Thus the Area of an arbitrary cross sectional square slice is:

# A_("slice") = l^2 #

# " " = (2sqrt(4-4x^2))^2 #

# " " = (4)(4-4x^2) #

# " " = 16-16x^2#

Finally, the volume of the entire solid is the sum of those arbitrary cross sectional slices over the elliptical base:

# V = sum_("ellipse") lim_(delta x rarr 0) A_("slice") delta x #

# \ \ \ = int_(-1)^(1) 16-16x^2 dx #

# \ \ \ = [16x-(16x^3)/3]_(-1)^(1) #

# \ \ \ = (16-16/3) - (-16+16/3 ) #

# \ \ \ = 16-16/3 +16-16/3 #

# \ \ \ = 64/3 #

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To calculate the volume of the solid with a base given by the ellipse equation (4x^2 + y^2 = 4) and vertical cross sections that are squares, integrate the area of a square cross section over the range of (x) values that cover the base of the solid. The area of each square cross section can be expressed as the side length of the square squared.

First, express (y) in terms of (x) from the equation of the ellipse. Then, find the limits of integration for (x). Since the ellipse is symmetric about the y-axis, we only need to consider the positive part of the ellipse. These limits correspond to the values of (x) where the ellipse intersects the y-axis.

Once you have the expression for (y) in terms of (x) and the limits of integration, integrate the square of the side length of the square cross section with respect to (x) over the determined range. Finally, evaluate the integral to find the volume of the solid.

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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.

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