Question

Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 4x and y = x2/4 . Find V by slicing.

Answer 1

Answer:

Step-by-step explanation:

Consider the graphs of the $y = 4x$  and  $y = \frac{x^{2} }{4}$.

By equating the expressions, the intersection points of the graphs can be found and in this way delimit the area that will rotate around the Y axis.

$4x = \frac{x^{2} }{4} \\   x^{2}  = 16x \\ x^{2}  - 16x = 0 \\   x(x-16) = 0$ then $x=0$  o  $x=16$. Therefore the integration limits are:

$y = 4(0) = 0$  and  $y = 4(16) = 64$

The inverse functions are given by:

$x = 2 \sqrt{y}$  and  $x = \frac{y}{4}$. Then

The volume of the solid of revolution is given by:

$\int\limits^{64}_ {0} \, [2\sqrt{y} - \frac{y}{4}]^{2}  dy = \int\limits^{64}_ {0} \, [4y - y^{3/2} + \frac{y^{2}}{16} ]\  dy = [2y^{2} - \frac{2}{5}y^{5/2} + \frac{y^{3}}{48} ]\limits^{64}_ {0} = 546.133 u^{2}$

Answer 2

The area under the curve of the volume of solid of revolution is 541.6.

Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 4x and $\rm y = \dfrac{x^2}{4}$

What is the area under the curve?

The area under the curve is the area bounded by the graph and is determined by finding the area between two graphs was found by doing a definite integral between two points.

To find the area bound by both graphs first we have to find the point of intersection.

$\rm 4x = \dfrac{x^2}{4}\\\\4 \times 4x = x^2\\\\ 16x=x^2\\\\x^2-16x=0\\\\x(x-16)=0\\\\x=0 \\\\x-16=0 \ \ x=16$

When the value of x is 16 then the value of y is;

$\rm y = 4x\\\\y = 4 \times 16\\\\y = 64$

When the value of x is 0 then the value of y is;

$\rm y = 4x\\\\y = 4 \times 0\\\\y = 0$

Now find the value of the given graph in the form of x;

$\rm y = 4x\\\\x = \dfrac{y}{4}\\\\And \\\\y=\dfrac{x^2}{4}\\\\4y= x^2\\\\x = 2\sqrt{y}$

Therefore,

The area under the curve of the volume of solid is;

$\rm Area = \int\limits^{64}_0 ({2\sqrt{y}-\dfrac{y}{4}}) \, dy\\\\Area = \int\limits^{64}_0 ({2\sqrt{y}) \, dy\\- \int\limits^{64}_0 (\dfrac{y}{4}}) \, dy\\\\ Area = \left [2y^2-\dfrac{2}{5}y^{\frac{5}{2}} \right ]^{64}_0 - \left [ \dfrac{-y^3}{48} \right ]^{64}_0\\\\Area = 2(64)^2-\dfrac{2}{5}64^{\frac{5}{2}}+\dfrac{64^3}{48}\\\\Area =8192-13107.2+5461.33\\\\Area = 546.1$

Hence, The area under the curve of the volume of solid of revolution is 541.6.

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