Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola and a Line around the 𝑥-Axis | Nagwa Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola and a Line around the 𝑥-Axis | Nagwa

Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola and a Line around the 𝑥-Axis Mathematics • Third Year of Secondary School

Find the volume of the solid generated by revolving the region bounded by the curve 𝑦² = 1 − 𝑥 and the straight line 𝑥 = −4 a complete revolution about the 𝑥-axis.

02:40

Video Transcript

Find the volume of the solid generated by revolving the region bounded by the curve 𝑦 squared equals one minus 𝑥 and the straight line 𝑥 equals negative four a complete revolution about the 𝑥-axis.

Let’s begin by sketching the curve 𝑦 squared equals one minus 𝑥. We can rearrange this equation for 𝑥. And we see that 𝑥 is a negative quadratic function of 𝑦. So, we will have an inverted parabola centered on the 𝑥-axis. At 𝑥 equals zero, 𝑦 squared must be equal to one. So, the 𝑦-intercepts are at plus or minus one. At 𝑦 equals zero, 𝑥 is equal to one. So, the 𝑥 intercept is one. Therefore, the curve of 𝑦 squared equals one minus 𝑥 looks like this. The line 𝑥 equals negative four is this vertical line here. Therefore, the region enclosed by the curve 𝑦 squared equals one minus 𝑥 and the line 𝑥 equals negative four is the shaded region.

The question asks us to find the volume of the solid generated by revolving this region about the 𝑥-axis. Recall that to find the volume of revolution about the 𝑥-axis of a region enclosed by a curve 𝑦 equals 𝑓 of 𝑥 and two vertical lines 𝑥 equals 𝑎 and 𝑥 equals 𝑏, we take 𝜋 times the integral between the limits of 𝑎 and 𝑏 of 𝑓 of 𝑥 all squared with respect to 𝑥. In this question, we are in fact given 𝑓 of 𝑥 squared explicitly, since 𝑦 squared is equal to one minus 𝑥. The limits 𝑎 and 𝑏 will be 𝑥 equals negative four and the upper 𝑥 limit of the region, which is the turning point of the parabola.

Since the parabola is symmetrical about the 𝑥-axis, this is just the 𝑥-intercept at 𝑥 equals one. Therefore, the volume of revolution of this region about the 𝑥-axis is given by 𝜋 times the integral between negative four and one of one minus 𝑥 with respect to 𝑥. Integrating, we obtain 𝜋 times 𝑥 minus 𝑥 squared over two evaluated between negative four and one. Evaluating, this gives us 𝜋 times one minus one squared over two minus negative four minus negative four all squared over two. Simplifying, this gives 𝜋 times one-half minus negative 12. Expanding the parentheses and simplifying gives the volume of the solid, 25𝜋 over two cubic units.

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