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Let m greater than 1 the vilume of the solid generated by revolving the region between the y axis and the curve xy=4 1 greater than or equal to y less than or equal to m about y axis is 15 pi the value of m is?
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Let m greater than 1 the vilume of the solid generated by revolving th...
**Solution:**

To find the value of *m* such that the volume of the solid generated by revolving the region between the y-axis and the curve *xy=4* about the y-axis is *15π*, we need to follow these steps:

**Step 1: Determine the limits of integration**

To find the limits of integration, we need to find the values of *y* at which the curve intersects the y-axis. From the equation *xy=4*, we can substitute *x=0* to get *y=0*. So, the lower limit of integration is *y=0*.

To find the upper limit of integration, we need to find the value of *y* such that the curve intersects the line *y=m*. We can substitute *y=m* into the equation *xy=4* to get *xm=4*. Solving for *x*, we have *x=4/m*. So, the upper limit of integration is *y=m*.

**Step 2: Set up the integral**

The volume of the solid generated by revolving the region between the y-axis and the curve *xy=4* about the y-axis can be calculated using the formula for the volume of a solid of revolution:

*V = ∫[a,b] π(f(y))^2 dy*

where *a* and *b* are the lower and upper limits of integration, respectively, and *f(y)* represents the function that gives the radius of the cross-section at each value of *y*.

In this case, the radius of the cross-section is given by *x*. Since *xy=4*, we can solve for *x* to get *x=4/y*. Therefore, the volume can be calculated as:

*V = ∫[0,m] π(4/y)^2 dy*

**Step 3: Evaluate the integral**

To evaluate the integral, we can simplify the expression inside the integral:

*V = ∫[0,m] π(16/y^2) dy*

Simplifying further:

*V = π ∫[0,m] (16/y^2) dy*

*V = π [-16/y] [0,m]*

*V = π (-16/m) + 16/0*

Since *m* is greater than 1, the volume of the solid generated by revolving the region between the y-axis and the curve *xy=4* about the y-axis is *15π*, we can set up the equation:

*15π = π (-16/m) + 16/0*

Simplifying further:

*15 = -16/m + 16/0*

Since *16/0* is undefined, the equation is not solvable. Therefore, there is no value of *m* that satisfies the given conditions.

Hence, there is no solution to this problem.

Note: It is important to note that the equation *16/0* is undefined because division by zero is not defined in mathematics.
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Let m greater than 1 the vilume of the solid generated by revolving the region between the y axis and the curve xy=4 1 greater than or equal to y less than or equal to m about y axis is 15 pi the value of m is?
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Let m greater than 1 the vilume of the solid generated by revolving the region between the y axis and the curve xy=4 1 greater than or equal to y less than or equal to m about y axis is 15 pi the value of m is? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Let m greater than 1 the vilume of the solid generated by revolving the region between the y axis and the curve xy=4 1 greater than or equal to y less than or equal to m about y axis is 15 pi the value of m is? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let m greater than 1 the vilume of the solid generated by revolving the region between the y axis and the curve xy=4 1 greater than or equal to y less than or equal to m about y axis is 15 pi the value of m is?.
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