I want the part c visualization ---**Textual Information:** The diagram shows the curve $y = 2x^2 - 18$ and straight line AB that is the tangent to the curve at point Q. Rajah menunjukkan lengkung $y = 2x^2 - 18$ dan garis lurus AB yang merupakan tangen kepada lengkung pada titik Q. Given that the gradient of straight line AB is 4. Diberi bahawa kecerunan garis lurus AB ialah 4. a) Find the coordinates of point Q. [2 marks] Cari koordinat titik Q. [2 markah] b) Calculate the area of shaded region. [5 marks] Hitung luas rantau berlorek. [5 markah] c) When the area bounded by the curve, x-axis and straight line $y = h$ is rotated through $180^\circ$ about the y-axis, the volume generated is $65\pi$ unit³. Find the value of $h$. [3 marks] Apabila rantau yang dibatasi oleh lengkung, paksi-x dan garis lurus $y = h$ diputarkan melalui $180^\circ$ pada paksi-y, isi padu kisaran ialah $65\pi$ unit³. Cari nilai $h$. [3 markah] **Chart/Diagram Description:** * **Type:** Coordinate plane with a curve and a tangent line. * **Coordinate Axes:** X-axis and Y-axis, labelled 'x' and 'y' respectively, intersecting at the origin O. * **Curve:** A parabola opening upwards, described by the equation $y = 2x^2 - 18$. It intersects the negative y-axis. * **Straight Line AB:** A straight line tangent to the curve at point Q. * **Points:** * O: Origin (0,0). * Q: Point of tangency on the curve and the line AB. * A: Point on the line AB, located on the negative y-axis. * B: Point on the line AB, located on the positive x-axis. * **Labels:** * 'O', 'Q', 'A', 'B', 'x', 'y'. * 'h': A mark on the positive y-axis above O. * '5': A numerical label on the positive x-axis, near point B. * **Shaded Region:** An area bounded by the curve $y = 2x^2 - 18$, the straight line AB, and the x-axis. This region is located in the first quadrant (x > 0, y > 0). It appears to be bounded by the x-axis from below, the curve on the left, and the line AB on the right and above. * **Relative Position and Direction:** Line AB passes through the first, fourth, and negative y-axis regions. Point Q is in the first quadrant or on the positive x-axis part of the boundary. Point B is on the positive x-axis. Point A is on the negative y-axis. The shaded region is above the x-axis, enclosed by the curve and the line AB. The mark 'h' is on the positive y-axis, below the shaded region.

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Let's solve part c of this problem. We need to find the value of h when the region bounded by the parabola y equals 2x squared minus 18, the x-axis, and the horizontal line y equals h is rotated 180 degrees about the y-axis to generate a volume of 65π cubic units. Here we see the coordinate system with the parabola intersecting the y-axis at negative 18, the tangent line AB, and the horizontal line y equals h that will define our region of interest. Now let's clearly define the region that will be rotated. The shaded region is bounded by three elements: the parabola y equals 2x squared minus 18, the x-axis where y equals 0, and the horizontal line y equals h. When this region is rotated 180 degrees about the y-axis, it creates a solid of revolution. The volume of this solid can be calculated using the disk method with the formula V equals π times the integral from negative 18 to h of x squared dy. To set up the volume integral, we need to express x in terms of y. Starting with y equals 2x squared minus 18, we solve for x squared to get x squared equals y plus 18 over 2. Since we're integrating with respect to y, our volume formula becomes V equals π times the integral from negative 18 to h of x squared dy, which substitutes to π times the integral from negative 18 to h of y plus 18 over 2 dy. The limits of integration are from y equals negative 18, where the curve intersects the y-axis, up to y equals h. Now let's evaluate the integral step by step. Starting with V equals π times the integral from negative 18 to h of y plus 18 over 2 dy, we factor out the constant π over 2. The antiderivative of y plus 18 is y plus 18 squared over 2. Applying the fundamental theorem of calculus, we evaluate from negative 18 to h. At y equals negative 18, we get zero, and at y equals h, we get h plus 18 squared. This gives us the final formula: V equals π times h plus 18 squared over 4. Finally, let's solve for h by setting our volume formula equal to 65π. We have π times h plus 18 squared over 4 equals 65π. Dividing both sides by π gives us h plus 18 squared over 4 equals 65. Multiplying by 4, we get h plus 18 squared equals 260. Taking the square root, h plus 18 equals 2 times the square root of 65. Therefore, h equals 2 times the square root of 65 minus 18, which is approximately negative 1.88. We can verify this answer by substituting back: the volume equals π times 260 over 4, which equals 65π, confirming our solution is correct.